Once again, Boris needs the help of Anton in creating a task. This time Anton needs to solve the following problem:

There are two arrays of integers a and b of length n. It turned out that array a contains only elements from the set \{-1, 0, 1\}.

Anton can perform the following sequence of operations any number of times:

  1. Choose any pair of indexes (i, j) such that 1 ≤ i < j ≤ n. It is possible to choose the same pair (i, j) more than once. 
  2. Add a_i to a_j. In other words, j-th element of the array becomes equal to a_i + a_j. 



For example, if you are given array [1, -1, 0], you can transform it only to [1, -1, -1], [1, 0, 0] and [1, -1, 1] by one operation.

Anton wants to predict if it is possible to apply some number (zero or more) of these operations to the array a so that it becomes equal to array b. Can you help him?

Input

Each test contains multiple test cases. 

The first line contains the number of test cases t (1 ≤ t ≤ 10000). The description of the test cases follows.

The first line of each test case contains a single integer n (1 ≤ n ≤ 10^5) — the length of arrays.

The second line of each test case contains n integers a_1, a_2, ..., a_n (-1 ≤ a_i ≤ 1) — elements of array a. There can be duplicates among elements.

The third line of each test case contains n integers b_1, b_2, ..., b_n (-10^9 ≤ b_i ≤ 10^9) — elements of array b. There can be duplicates among elements.

It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.

Output

For each test case, output one line containing "YES" if it's possible to make arrays a and b equal by performing the described operations, or "NO" if it's impossible.

You can print each letter in any case (upper or lower).

Example

Input


5
3
1 -1 0
1 1 -2
3
0 1 1
0 2 2
2
1 0
1 41
2
-1 0
-1 -41
5
0 1 -1 1 -1
1 1 -1 1 -1


Output


YES
NO
YES
YES
NO

Note

In the first test-case we can choose (i, j)=(2, 3) twice and after that choose (i, j)=(1, 2) twice too. These operations will transform [1, -1, 0] → [1, -1, -2] → [1, 1, -2]

In the second test case we can't make equal numbers on the second position.

In the third test case we can choose (i, j)=(1, 2) 41 times. The same about the fourth test case.

In the last lest case, it is impossible to make array a equal to the array b.
import java.io.BufferedReader;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashMap;
import java.util.HashSet;
import java.util.LinkedList;
import java.util.PriorityQueue;
import java.util.Queue;
import java.util.StringTokenizer;
import java.util.TreeSet;

public class Main {
	static ArrayList<Edge> adj[];
	static PrintWriter out = new PrintWriter(System.out);
	public static long mod;

	static int [][]notmemo;
	static int k;
	static int a[];
	static int b[];
	static int m;
	static char c[];


	static class Pair implements Comparable<Pair>{
    	int x;
    	int y;
    	public Pair(int a,int b) {
		x=a;
		y=b;
		
    	}
		@Override
		public int compareTo(Pair o) {
			if(x!=o.x)
			return x-o.x;
			return y-o.y;
		}
    	
    }
    static Pair s1[];
    static int s[];
    static ArrayList<Pair> adjlist[];
	private static HashSet<Integer> set1[];
	public static void main(String args[]) throws IOException {
		Scanner sc = new Scanner(System.in);
		int t=sc.nextInt();
		while(t-->0) {
			int n=sc.nextInt();
		int a[]=new int[n];
	int b[]=new int[n];
	for (int i = 0; i < b.length; i++) {
		a[i]=sc.nextInt();
	}
	for (int i = 0; i < b.length; i++) {
		b[i]=sc.nextInt();
	}
	HashSet<Integer> set=new HashSet<>();
	boolean f=true;
	for (int i = 0; i < b.length; i++) {
		if(b[i]>a[i]&&!set.contains(1)) {
			f=false;
		}
		else if(a[i]>b[i]&&!set.contains(-1)) {
			f=false;
		}
		set.add(a[i]);
	}
	if(!f)
		out.println("NO");
	else
		out.println("YES");
		}
		out.flush();
	}
	public static int countvowels() {
		int count=0;
		for (int i = 0; i < c.length; i++) {
			if(c[i]=='o'||c[i]=='e'||c[i]=='a'||c[i]=='i'||c[i]=='u') {
				count++;
			}
		}
		return count;
	}
	public static  char lastvowel() {
		for (int i = c.length-1; i>=0; i--) {
			if(c[i]=='o'||c[i]=='e'||c[i]=='a'||c[i]=='i'||c[i]=='u') {
		return c[i];
			}
		}
		return 'a';
	}
	static class pp{
		String ss;
		String mm;
		public pp(String a,String b) {
		ss=a;
		mm=b;
		}
	}
	static class Tri implements Comparable<Tri>{
		String s;
		int numofvow;
		char lastchar;
	public	Tri(String m,int a,char c1){
			s=m;
			numofvow=a;
			lastchar=c1;
			
		}
	public int compareTo(Tri t) {
		if (numofvow == t.numofvow)
			return lastchar - t.lastchar;
		return numofvow - t.numofvow;
	}
	
	}
	static boolean dfs(int u, int cost) {
		boolean f=false;
		vis[u]=true;
		for(Edge p:adj[u]) {
			if(!vis[p.node]) {
				f|=dfs(p.node,(int) p.cost);
			}
		}
		if (cost == 2&&!f) {
				set.add(u + 1);
				return true;
		}
			return f;
		
	}
	static int min(int i,int last) {
		if(i==a.length)
			return 0;
		int take=0;
		if(gcd(a[i],last)>1||i==0) {
			take=Math.max(take,1+min(i+1,a[i]));
		}
	 take=Math.max(take,min(i+1,last));
		
		return b[i]=take;
	}
		
	private static boolean check(String s) {
		char c1[]=s.toCharArray();
		for (int i = 0; i < c1.length; i++) {
			if(c[i]>c1[i]) {
				return true;
			}
			 if(c[i]<c1[i])
				return false;
		}
		return true;
	}

	static long dijkstra(int S, int T)	//O(E log E)
	{
		long[] dist = new long[V];
		Arrays.fill(dist, (int) 1e9);
		PriorityQueue<Edge> pq = new PriorityQueue<Edge>();
		dist[S] = 0;
		pq.add(new Edge(S, 0));						//may add more in case of MSSP (Mult-Source)
		while(!pq.isEmpty())
		{
			Edge cur = pq.remove();
			if(cur.node == T)						//remove if all computations are needed
				return dist[T];
			if(cur.cost > dist[cur.node])			//lazy deletion
				continue;
			for(Edge nxt: adj[cur.node])
				if(cur.cost + nxt.cost < dist[nxt.node])
					pq.add(new Edge(nxt.node, dist[nxt.node] = cur.cost + nxt.cost ));
		}
		return -1;
	}
	static class Edge implements Comparable<Edge>
	{
		int node;long cost;
		
		Edge(int a, long b) { node = a;	cost = b; }
		
		public int compareTo(Edge e){ return Long.compare(cost,e.cost);	}
	}
	
    static int dp(int idx,int state) {
      if(idx==n) {
    	  return 0;
      }
      if(notmemo[idx][state]!=-1) {
    	  return notmemo[idx][state];
      }
     int take=0; int leave=0;int leaveall=0;
      if(a[idx]==1&&(state==1||state==0)) {
    	take=1+dp(idx+1,1);  
      }
      if(a[idx]==0&&state==0) {
    	 leave=1+dp(idx+1,state); 
      }
      leaveall=dp(idx+1,state);
      
      return notmemo[idx][state]=Math.max(take,Math.max(leaveall,leave));
    }
        static void sieve(int N)	// O(N log log N) 
	{
		isComposite = new int[N+1];					
		isComposite[0] = isComposite[1] = 1;			// 0 indicates a prime number
		primes = new ArrayList<Integer>();
 
		for (int i = 2; i <= N; ++i) 					//can loop till i*i <= N if primes array is not needed O(N log log sqrt(N)) 
			if (isComposite[i] == 0) 					//can loop in 2 and odd integers for slightly better performance
			{
				primes.add(i);
				if(1l * i * i <= N)
					for (int j = i * i; j <= N; j += i)	// j = i * 2 will not affect performance too much, may alter in modified sieve
						isComposite[j] = 1;
			}   
	}
    static TreeSet<Integer> factors;
    static ArrayList<Integer> primeFactors(int N)		// O(sqrt(N) / ln sqrt(N))
	{
		ArrayList<Integer> factors = new ArrayList<Integer>();		//take abs(N) in case of -ve integers
		int idx = 0, p = primes.get(idx);

		while(1l*p * p <= N)
		{
			while(N % p == 0) { factors.add(p); N /= p; }
			p = primes.get(++idx);
		}

		if(N != 1)						// last prime factor may be > sqrt(N)
			factors.add(N);				// for integers whose largest prime factor has a power of 1
		return factors;
	}
    
    static String y;
    	static int nomnom[];
	static long fac[];
	
	static class book implements Comparable<book> {
		int idx;
		long score;

		public book(int i, long s) {
			idx = i;
			score = s;
		}

		public int compareTo(book o) {
			return (int) (o.score - score);
		}
	}

	static class library implements Comparable<library> {
		int numofbooks;
		int signup;
		int shiprate;
		int idx;

		public library(int a, int b, int c, int idx) {
			numofbooks = a;
			signup = b;
			shiprate = c;
			this.idx = idx;
		}

		@Override
		public int compareTo(library o) {
		if(signup==o.signup) {
			  return o.numofbooks-numofbooks;
		}
		return signup - o.signup;
		
		}
	}





	static boolean isOn(int S, int j) {
		return (S & 1 << j) != 0;
	}


	static boolean f = true;

	static class SegmentTree { // 1-based DS, OOP

		int N; // the number of elements in the array as a power of 2 (i.e. after padding)
		int[] array, sTree, lazy;

		SegmentTree(int[] in) {
			array = in;
			N = in.length - 1;
			sTree = new int[N << 1]; // no. of nodes = 2*N - 1, we add one to cross out index zero
			lazy = new int[N << 1];
			//build(1, 1, N);
		}

		void build(int node, int b, int e) // O(n)
		{
			if (b == e)
				sTree[node] = array[b];
			else {
				int mid = b + e >> 1;
				build(node << 1, b, mid);
				build(node << 1 | 1, mid + 1, e);
				sTree[node] = sTree[node << 1] + sTree[node << 1 | 1];
			}
		}

		void update_point(int index, int val) // O(log n)
		{
			index += N - 1;				
			sTree[index] = val;			
			while(index>1)				
			{
				index >>= 1;
				sTree[index] = Math.max(sTree[index<<1] ,sTree[index<<1|1]);		
			}
		}

		void update_range(int i, int j, int val) // O(log n)
		{
			update_range(1, 1, N, i, j, val);
		}

		void update_range(int node, int b, int e, int i, int j, int val) {
			if (i > e || j < b)
				return;
			if (b >= i && e <= j) {
				sTree[node] += (e - b + 1) * val;
				lazy[node] += val;
			} else {
				int mid = b + e >> 1;
				propagate(node, b, mid, e);
				update_range(node << 1, b, mid, i, j, val);
				update_range(node << 1 | 1, mid + 1, e, i, j, val);
				sTree[node] = sTree[node << 1] + sTree[node << 1 | 1];
			}

		}

		void propagate(int node, int b, int mid, int e) {
			lazy[node << 1] += lazy[node];
			lazy[node << 1 | 1] += lazy[node];
			sTree[node << 1] += (mid - b + 1) * lazy[node];
			sTree[node << 1 | 1] += (e - mid) * lazy[node];
			lazy[node] = 0;
		}

		int query(int i, int j) {
			return query(1, 1, N, i, j);
		}

		int query(int node, int b, int e, int i, int j) // O(log n)
		{
			if (i > e || j < b)
				return 0;
			if (b >= i && e <= j)
				return sTree[node];
			int mid = b + e >> 1;
			// propagate(node, b, mid, e);
			int q1 = query(node << 1, b, mid, i, j);
			int q2 = query(node << 1 | 1, mid + 1, e, i, j);
			return Math.max(q1,q2);

		}
	}

	static int memo[];


	static class UnionFind {
		int[] p, rank, setSize;
		int numSets;
		int max[];

		public UnionFind(int N) {
			p = new int[numSets = N];
			rank = new int[N];
			setSize = new int[N];
			for (int i = 0; i < N; i++) {
				p[i] = i;
				setSize[i] = 1;
			}
		}

		public int findSet(int i) {
			return p[i] == i ? i : (p[i] = findSet(p[i]));
		}

		public boolean isSameSet(int i, int j) {
			return findSet(i) == findSet(j);
		}

		public void unionSet(int i, int j) {
			if (isSameSet(i, j))
				return;
			numSets--;
			int x = findSet(i), y = findSet(j);
			if (rank[x] > rank[y]) {
				p[y] = x;
				setSize[x] += setSize[y];


			} else {
				p[x] = y;
				setSize[y] += setSize[x];
				if (rank[x] == rank[y])
					rank[y]++;


			}
		}

		

		public int numDisjointSets() {
			return numSets;
		}

		public int sizeOfSet(int i) {
			return setSize[findSet(i)];
		}
	}

	/**
	 * private static void trace(int i, int time) { if(i==n) return;
	 * 
	 * 
	 * long ans=dp(i,time);
	 * if(time+a[i].t<a[i].burn&&(ans==dp(i+1,time+a[i].t)+a[i].cost)) {
	 * 
	 * trace(i+1, time+a[i].t);
	 * 
	 * l1.add(a[i].idx); return; } trace(i+1,time);
	 * 
	 * }
	 **/

	static class incpair implements Comparable<incpair> {
		int a;
		long b;
		int idx;

		incpair(int a, long dirg, int i) {
			this.a = a;
			b = dirg;
			idx = i;
		}

		public int compareTo(incpair e) {
			return (int) (b - e.b);
		}
	}

	static class decpair implements Comparable<decpair> {
		int a;
		long b;
		int idx;

		decpair(int a, long dirg, int i) {
			this.a = a;
			b = dirg;
			idx = i;
		}

		public int compareTo(decpair e) {
			return (int) (e.b - b);
		}
	}

	static long allpowers[];

	static class Quad implements Comparable<Quad> {
		int u;
		int v;
		char state;
		int turns;

		public Quad(int i, int j, char c, int k) {
			u = i;
			v = j;
			state = c;
			turns = k;
		}

		public int compareTo(Quad e) {
			return (int) (turns - e.turns);
		}

	}

	static long dirg[][];
	static Edge[] driver;

	static int n;

	static long manhatandistance(long x, long x2, long y, long y2) {
		return Math.abs(x - x2) + Math.abs(y - y2);
	}

	static long fib[];

	static long fib(int n) {
		if (n == 1 || n == 0) {
			return 1;
		}
		if (fib[n] != -1) {
			return fib[n];
		} else
			return fib[n] = ((fib(n - 2) % mod + fib(n - 1) % mod) % mod);
	}

	static class Point  implements Comparable<Point>{
		long x, y;

		Point(long counth, long counts) {
			x = counth;
			y = counts;
		}

		@Override
		public int compareTo(Point p )
		{
			return Long.compare(p.y*1l*x, p.x*1l*y);
		}
		


	}

	static long[][] comb;

	static class Triple implements Comparable<Triple> {

		int l;
		int r;
		long cost;
		int idx;

		public Triple(int a, int b, long l1, int l2) {
			l = a;
			r = b;
			cost = l1;
			idx = l2;
		}

		public int compareTo(Triple x) {
			if (l != x.l || idx == x.idx)
				return l - x.l;
			return -idx;
		}

	}

	static TreeSet<Long> primeFactors(long N) // O(sqrt(N) / ln sqrt(N))
	{
		TreeSet<Long> factors = new TreeSet<Long>(); // take abs(N) in case of -ve integers
		int idx = 0, p = primes.get(idx);

		while (p * p <= N) {
			while (N % p == 0) {
				factors.add((long) p);
				N /= p;
			}
			if (primes.size() > idx + 1)
				p = primes.get(++idx);
			else
				break;
		}

		if (N != 1) // last prime factor may be > sqrt(N)
			factors.add(N); // for integers whose largest prime factor has a power of 1
		return factors;
	}

	static boolean visited[];

	/**
	 * static int bfs(int s) { Queue<Integer> q = new LinkedList<Integer>();
	 * q.add(s); int count=0; int maxcost=0; int dist[]=new int[n]; dist[s]=0;
	 * while(!q.isEmpty()) {
	 * 
	 * int u = q.remove(); if(dist[u]==k) { break; } for(Pair v: adj[u]) {
	 * maxcost=Math.max(maxcost, v.cost);
	 * 
	 * 
	 * 
	 * if(!visited[v.v]) {
	 * 
	 * visited[v.v]=true; q.add(v.v); dist[v.v]=dist[u]+1; maxcost=Math.max(maxcost,
	 * v.cost); } }
	 * 
	 * } return maxcost; }
	 **/
	public static boolean FindAllElements(int n, int k) {
		int sum = k;
		int[] A = new int[k];
		Arrays.fill(A, 0, k, 1);

		for (int i = k - 1; i >= 0; --i) {

			while (sum + A[i] <= n) {

				sum += A[i];
				A[i] *= 2;
			}
		}
		if (sum == n) {
			return true;
		} else
			return false;
	}

	static boolean[] vis2;

	static boolean f2 = false;

	static long[][] matMul(long[][] a2, long[][] b, int p, int q, int r) // C(p x r) = A(p x q) x (q x r) -- O(p x q x
																			// r)
	{
		long[][] C = new long[p][r];
		for (int i = 0; i < p; ++i) {
			for (int j = 0; j < r; ++j) {
				for (int k = 0; k < q; ++k) {
					C[i][j] = (C[i][j] + (a2[i][k] % mod * b[k][j] % mod)) % mod;
					C[i][j] %= mod;
				}

			}
		}
		return C;
	}

	static int memo1[];

	static boolean vis[];
	static TreeSet<Integer> set = new TreeSet<Integer>();

	static long modPow(long ways, long count, long mod) // O(log e)
	{
		ways %= mod;
		long res = 1;
		while (count > 0) {
			if ((count & 1) == 1)
				res = (res * ways) % mod;
			ways = (ways * ways) % mod;
			count >>= 1;
		}
		return res % mod;
	}

	static long gcd(long ans, long b) {
		if (b == 0) {
			return ans;
		}
		return gcd(b, ans % b);
	}

	static int[] isComposite;
	static int[] valid;

	static ArrayList<Integer> primes;
	static ArrayList<Integer> l1;

	
	static TreeSet<Integer> primus = new TreeSet<Integer>();
	
	static void sieveLinear(int N)
	{
		int[] lp = new int[N + 1];								//lp[i] = least prime divisor of i
		for(int i = 2; i <= N; ++i)
		{
			if(lp[i] == 0)
			{
				primus.add(i);
				lp[i] = i;
			}
			int curLP = lp[i];
			for(int p: primus)
				if(p > curLP || p * i > N)
					break;
				else
					lp[p * i] = i;
		}
	}
	

	public static int[] schuffle(int[] p) {
		for (int i = 0; i < p.length; i++) {
			int x = (int) (Math.random() * p.length);
			int temp = p[x];
			p[x] = p[i];
			p[i] = temp;
		}
		return p;
	}

	static int V;
	static long INF = (long) 1E16;

	static class Edge2 {
		int node;
		long cost;
		long next;

		Edge2(int a, int c, Long long1) {
			node = a;
			cost = long1;
			next = c;
		}

	}

	static class Scanner {
		StringTokenizer st;
		BufferedReader br;

		public Scanner(InputStream system) {
			br = new BufferedReader(new InputStreamReader(system));
		}

		public Scanner(String file) throws Exception {
			br = new BufferedReader(new FileReader(file));
		}

		public String next() throws IOException {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}

		public String nextLine() throws IOException {
			return br.readLine();
		}

		public int nextInt() throws IOException {
			return Integer.parseInt(next());
		}

		public double nextDouble() throws IOException {
			return Double.parseDouble(next());
		}

		public char nextChar() throws IOException {
			return next().charAt(0);
		}

		public Long nextLong() throws IOException {
			return Long.parseLong(next());
		}

		public boolean ready() throws IOException {
			return br.ready();
		}

		public void waitForInput() throws InterruptedException {
			Thread.sleep(3000);
		}

		public int[] nxtArr(int n) throws IOException {
			int[] ans = new int[n];
			for (int i = 0; i < n; i++)
				ans[i] = nextInt();
			return ans;
		}
	}
}
