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 javax.print.DocFlavor;
import javax.swing.*;
import java.awt.image.BandedSampleModel;
import java.util.*;

public class Example {
    static long[][] dp;
    //  static    int[]  ar;
    static int mod = 1000000000 + 7;
    static long[] ans = new long[100000 + 1];


    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int  t=sc.nextInt();
        while(t>0){
            t--;
            int n =sc.nextInt();
            int[] arr= new int[n];
            for(int i=0;i<n;i++){
arr[i]=sc.nextInt();
            }
            int[]  b=new int[n];
            for(int i=0;i<n;i++){
                b[i]=sc.nextInt();
            }
            boolean ans=true;
            boolean p=false;
            boolean neg=false;
            for(int i=0;i<n;i++){

                if(b[i]>arr[i]){
                    if(!p){
                        ans=false;
                        break;
                    }
                }else if(b[i]<arr[i]){
                    if(!neg){
                        ans=false;
                        break;
                    }
                }
                if(arr[i]==-1){
                    neg=true;
                }else if(arr[i]==1){
                    p=true;
                }


            }
            if(ans)
                System.out.println("YES");
            else
                System.out.println("NO");

        }





    }

    private static boolean isPrime(long a) {
        if (a == 1)
            return true;

        for (int i = 2; i * i <= a; i++) {
            if (a % i == 0)
                return false;

        }
        return true;
    }

    private static int gcd(int aa, int bb) {
        if (bb == 0)
            return aa;
        return gcd(bb, aa % bb);

    }

    private static int lowerBound(int[] ar, int a) {
        int res = 0;
        int s = 1;
        int e = ar.length - 1;
        while (s <= e) {
            int mid = (s) + (e - s) / 2;
            if (ar[mid] <= a) {
                res = mid;
                s = mid + 1;

            } else {
                e = mid - 1;
            }
        }
        return res;
    }

    public static int getXORSum(int[] arr1, int[] arr2) {

        long c = 0;
        int n = arr1.length;
        int m = arr2.length;
        if (n == 1 && m == 1) {
            return arr1[0] & arr2[0];
        }
        int aa = n + m;
        for (int i = 0; i < 32; i++) {

            int z = 0;
            int o = 0;
            int cc = 0;
            for (int a : arr1) {


                int x = (a & (1 << i));
                if (x == 0)
                    z++;
                else
                    o++;


            }
            for (int a : arr2) {
                int x = (a & (1 << i));
                if (x == 0)
                    z++;
                else
                    o++;
            }

            System.out.println(z + " " + o);

            if (z != aa && o != aa) {
                c = c + (1 << i);

            }


        }
        return (int) c;


    }


}





  	 			  	  	 	 	 		  	  						