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.*;
public class Main
{
    public static void main(String args[])throws Exception
    {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        PrintWriter pw=new PrintWriter(System.out);
        int t=Integer.parseInt(br.readLine());
        for(int x=0;x<t;x++)
        {
            int n=Integer.parseInt(br.readLine());
            String str[]=br.readLine().split(" ");
            int cntr[][]=new int[n][3];
            int arr1[]=new int[n];
            int arr2[]=new int[n];
            arr1[0]=Integer.parseInt(str[0]);
            for(int i=1;i<n;i++)
            {
                arr1[i]=Integer.parseInt(str[i]);
                cntr[i][0]+=cntr[i-1][0];
                cntr[i][1]+=cntr[i-1][1];
                cntr[i][2]+=cntr[i-1][2];
                if(arr1[i-1]==0)
                cntr[i][0]++;
                else if(arr1[i-1]==1)
                cntr[i][1]++;
                if(arr1[i-1]==-1)
                cntr[i][2]++;
            }
            str=br.readLine().split(" ");
            for(int i=0;i<n;i++)
            arr2[i]=Integer.parseInt(str[i]);
            boolean b=true;
            for(int i=n-1;i>=0;i--)
            {
                if(arr2[i]>arr1[i]&&cntr[i][1]==0)
                {
                    b=false;
                    break;
                }
                if(arr2[i]<arr1[i]&&cntr[i][2]==0)
                {
                    b=false;
                    break;
                }
            }
            if(b)
            pw.println("YES");
            else
            pw.println("NO");
        }
        pw.flush();
        pw.close();
    }
}