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.util.*;
import java.io.*;
import java.math.BigInteger;
import java.text.*;
public class Main {
    static long mod = (long)1e9 + 7;
    static long mod1 = 998244353;
    static boolean fileIO = false;
    static FastScanner f;
    static PrintWriter pw = new PrintWriter(System.out);
    static Scanner S = new Scanner(System.in);
    public static void main(String[] args)throws IOException {
        init();
        int t = f.nextInt();
        while(t --> 0) {
            int n = f.nextInt();
            int a[] = inpint(n);
            int b[] = inpint(n);
            if(a[0] != b[0]) {pn("NO"); continue;}
            else {
                int min = a[0];
                int max = a[0];
                int flag = 0;
                for(int i = 1; i < n; ++i) {
                    if(a[i] != b[i]) {
                        if(a[i] < b[i]) {if(max <= 0) {flag = 1; break;}}
                        else {if(min >= 0) {flag = 1; break;}}
                    }
                    min = Math.min(a[i] , min); max = Math.max(max , a[i]);
                }
                if(flag == 1) pn("NO"); else pn("YES");
            }
        }
        pw.flush();
        pw.close();  
    }
        
/******************************END OF MAIN PROGRAM*******************************************/
    public static void init()throws IOException{if(fileIO){f=new FastScanner("");}else{f=new FastScanner(System.in);}}
    public static class FastScanner {
        BufferedReader br;StringTokenizer st;
        FastScanner(InputStream stream){try{br=new BufferedReader(new InputStreamReader(stream));}catch(Exception e){e.printStackTrace();}}
        FastScanner(String str){try{br=new BufferedReader(new FileReader("!a.txt"));}catch(Exception e){e.printStackTrace();}}
        String next(){while(st==null||!st.hasMoreTokens()){try{st=new StringTokenizer(br.readLine());}catch(IOException e){e.printStackTrace();}}return st.nextToken();}
        String nextLine()throws IOException{return br.readLine();}int nextInt(){return Integer.parseInt(next());}long nextLong(){return Long.parseLong(next());}double nextDouble(){return Double.parseDouble(next());}
    }
    public static void pn(Object o){pw.println(o);}
    public static void p(Object o){pw.print(o);}
    public static void pni(Object o){pw.println(o);pw.flush();}
    static class Point implements Comparator<Point>{int x;int y;Point(int x,int y){this.x=x;this.y=y;}Point(){}public int compare(Point a, Point b){if(a.x==b.x)return a.y-b.y;return a.x-b.x;}}
    static int gcd(int a,int b){if(b==0)return a;else{return gcd(b,a%b);}}
    static long gcd(long a,long b){if(b==0)return a;else{return gcd(b,a%b);}}
    static long pow(long a,long b){long res=1;while(b>0){if((b&1)==1)res=res*a;b>>=1;a=a*a;}return res;}
    static long mpow(long a,long b){long res=1;while(b>0){if((b&1)==1)res=((res%mod)*(a%mod))%mod;b>>=1;a=((a%mod)*(a%mod))%mod;}return res;}
    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(long i=5;i*i<=n;i=i+6)if(n%i==0||n%(i+2)==0)return false;return true;}
    static HashSet<Long> factors(long n){HashSet<Long> hs=new HashSet<Long>();for(long i=1;i<=(long)Math.sqrt(n);i++){if(n%i==0){hs.add(i);hs.add(n/i);}}return hs;}
    static int[] inpint(int n){int arr[]=new int[n];for(int i=0;i<n;i++){arr[i]=f.nextInt();}return arr;}
    static long[] inplong(int n){long arr[] = new long[n];for(int i=0;i<n;i++){arr[i]=f.nextLong();}return arr;}
    static HashSet<Integer> hsi(){return new HashSet<Integer>();}static HashSet<Long> hsl(){return new HashSet<Long>();}
    static ArrayList<Integer> ali(){return new ArrayList<Integer>();}static ArrayList<Long> all(){return new ArrayList<Long>();}
    static HashMap<Integer,Integer> hii(){return new HashMap<Integer,Integer>();}
    static HashMap<Integer,ArrayList<Integer>> g(){return new HashMap<Integer,ArrayList<Integer>>();}
    static boolean ise(int x){return ((x&1)==0);}static boolean ise(long x){return ((x&1)==0);}
}