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.
	/*	/ ﾌﾌ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ム
		/ )\⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀  Y
		(⠀⠀| ( ͡° ͜ʖ ͡°）⠀⌒(⠀ ノ
		(⠀ ﾉ⌒ Y ⌒ヽ-く __／
		| _⠀｡ノ| ノ｡ |/
		(⠀ー '_人`ー ﾉ
		⠀|\ ￣ _人'彡ﾉ
		⠀ )\⠀⠀ ｡⠀⠀ /
		⠀⠀(\⠀ #⠀ /
		⠀/⠀⠀⠀/ὣ====================D-
		/⠀⠀⠀/⠀ \ \⠀⠀\
		( (⠀)⠀⠀⠀⠀ ) ).⠀)
		(⠀⠀)⠀⠀⠀⠀⠀( | /
		|⠀ /⠀⠀⠀⠀⠀⠀ | /
		[_] ⠀⠀⠀⠀⠀[___]                     */
	// Main Code at the Bottom
	import java.util.*;
	import java.lang.*;
	import java.io.*;
	import java.math.BigInteger; 
	public class Main {
		//Fast IO class
	    static class FastReader {
	        BufferedReader br; 
	        StringTokenizer st; 
	        public FastReader() {
	        	boolean env=System.getProperty("ONLINE_JUDGE") != null;
	        	if(!env) {
	        		try {
						br=new BufferedReader(new FileReader("src\\input.txt"));
					} catch (FileNotFoundException e) {
						e.printStackTrace();
					}
	        	}
	        	else br = new BufferedReader(new InputStreamReader(System.in)); 
	        } 
	        String next() {
	            while (st == null || !st.hasMoreElements()) {
	                try {
	                    st = new StringTokenizer(br.readLine()); 
	                } 
	                catch (IOException  e) {
	                    e.printStackTrace(); 
	                } 
	            } 
	            return st.nextToken(); 
	        } 
	        int nextInt() {
	            return Integer.parseInt(next()); 
	        } 
	        long nextLong() {
	            return Long.parseLong(next()); 
	        } 
	        double nextDouble() {
	            return Double.parseDouble(next()); 
	        } 
	        String nextLine() {
	            String str = ""; 
	            try {
	                str = br.readLine(); 
	            } 
	            catch (IOException e) {
	                e.printStackTrace(); 
	            } 
	            return str; 
	        } 
	    }     
	    static long MOD=1000000000+7;
	    //Euclidean Algorithm
	    static long gcd(long A,long B){
	    	return (B==0)?A:gcd(B,A%B);
	    }
	    //Modular Exponentiation
	    static long fastExpo(long x,long n){
	        if(n==0) return 1;
	        if((n&1)==0) return fastExpo((x*x)%MOD,n/2)%MOD;
	        return ((x%MOD)*fastExpo((x*x)%MOD,(n-1)/2))%MOD;
	    }
	    //Modular Inverse
	    static long inverse(long x) {
	    	return fastExpo(x,MOD-2);
	    }
	    //Prime Number Algorithm
	    static boolean isPrime(long n){
	        if(n<=1) return false;
	        if(n<=3) return true;
	        if(n%2==0 || n%3==0) return false;
	        for(int i=5;i*i<=n;i+=6) if(n%i==0 || n%(i+2)==0) return false;
	        return true;
	    }
	    //Reverse an array
	    static void reverse(int arr[],int l,int r){
	    	while(l<r) {
	    		int tmp=arr[l];
	    		arr[l++]=arr[r];
	    		arr[r--]=tmp;
	    	}
	    }
	    //Print array
	    static void print1d(int arr[]) {
	    	out.println(Arrays.toString(arr));
	    }
	    static void print2d(int arr[][]) {
	    	for(int a[]: arr) out.println(Arrays.toString(a));
	    }
	    // Pair
	    static class pair{
	    	int x;
	    	double y;
	    	pair(int a,double b){
	    		this.x=a;
	    		this.y=b;
	    	}
	    	public boolean equals(Object obj) {
	    		if(obj == null || obj.getClass()!= this.getClass()) return false;
	            pair p = (pair) obj;
	            return (this.x==p.x && this.y==p.y);
	        }
	    	public int hashCode() {
	            return Objects.hash(x,y);
	        }
	    }
	    static FastReader sc=new FastReader();
	    static PrintWriter out=new PrintWriter(System.out);  
	    //Main function(The main code starts from here
	    public static void main (String[] args) throws java.lang.Exception {
	    	int test;
	    	test=1;
	    	test=sc.nextInt();
	    	while(test-->0){
	    		int n=sc.nextInt();
	    		long a[]=new long[n],b[]=new long[n];
	    		for(int i=0;i<n;i++) a[i]=sc.nextLong();
	    		for(int i=0;i<n;i++) b[i]=sc.nextLong();
	    		String ans="YES";
	    		int o=0,m=0;
	    		for(int i=0;i<n;i++) {
	    			if(a[i]==-1) m++;
	    			else if(a[i]==1) o++;
	    		}
	    		for(int i=n-1;i>=0;i--) {
	    			if(a[i]==-1) m--;
	    			else if(a[i]==1) o--;
	    			long diff=b[i]-a[i];
	    			if(diff>0 && o<=0) ans="NO";
	    			else if(diff<0 && m<=0) ans="NO";
	    		}
	    		out.println(ans);
	    	}
	        out.flush();
	        out.close();
	    }
	}