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.*;
import java.util.*;

public class B {
    public static void main(String[] args) throws IOException {
        BufferedReader inp = new BufferedReader(new InputStreamReader(System.in));
        BufferedWriter out = new BufferedWriter(new OutputStreamWriter(System.out));

        int testCase = Integer.parseInt(inp.readLine());

        for(int t=0;t<testCase;t++){

            int size = Integer.parseInt(inp.readLine());
            int[] given = new int[size];
            String[] s1 = inp.readLine().split(" ");
            for(int i=0;i<size;i++){
                given[i] = Integer.parseInt(s1[i]);
            }

            int[] copy = new int[size];
            s1 = inp.readLine().split(" ");
            for(int i=0;i<size;i++){
                copy[i] = Integer.parseInt(s1[i]);
            }

            boolean negOne = false;
            boolean zero = false;
            boolean one = false;

            if(given[0]==1){
                one = true;
            }
            if(given[0]==-1){
                negOne = true;
            }
            if(given[0]==0){
                zero = true;
            }

            boolean ans = true;

            if(given[0]!=copy[0]){
                ans = false;
            }

            for(int i=1;i<size;i++){

                if(given[i]<copy[i]){
                    if(!one){
                        ans =false;
                    }
                }
                if(given[i]>copy[i]){
                    if(!negOne){
                        ans = false;
                    }
                }

                if(given[i]==1){
                    one = true;
                }
                if(given[i]==-1){
                    negOne = true;
                }
                if(given[i]==0){
                    zero = true;
                }

            }

            if(ans){
                out.write("YES"+"\n");
            }
            else{
                out.write("NO"+"\n");
            }


        }
        out.flush();
    }
    static void merge(int arr[], int l, int m, int r)
    {
        // Find sizes of two subarrays to be merged
        int n1 = m - l + 1;
        int n2 = r - m;

        /* Create temp arrays */
        int L[] = new int [n1];
        int R[] = new int [n2];

        /*Copy data to temp arrays*/
        for (int i=0; i<n1; ++i)
            L[i] = arr[l + i];
        for (int j=0; j<n2; ++j)
            R[j] = arr[m + 1+ j];


        /* Merge the temp arrays */

        // Initial indexes of first and second subarrays
        int i = 0, j = 0;

        // Initial index of merged subarry array
        int k = l;
        while (i < n1 && j < n2)
        {
            if (L[i] <= R[j])
            {
                arr[k] = L[i];
                i++;
            }
            else
            {
                arr[k] = R[j];
                j++;
            }
            k++;
        }

        /* Copy remaining elements of L[] if any */
        while (i < n1)
        {
            arr[k] = L[i];
            i++;
            k++;
        }

        /* Copy remaining elements of R[] if any */
        while (j < n2)
        {
            arr[k] = R[j];
            j++;
            k++;
        }
    }

    // Main function that sorts arr[l..r] using
    // merge()
    static void sort(int arr[], int l, int r)
    {
        if (l < r)
        {
            // Find the middle point
            int m = (l+r)/2;

            // Sort first and second halves
            sort(arr, l, m);
            sort(arr , m+1, r);

            // Merge the sorted halves
            merge(arr, l, m, r);
        }
    }
}
