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.
/*package whatever //do not write package name here */

import java.io.*;
import java.util.*;

public class GFG {
    public static void fun(int a[],int b[],int n)
    {
        int p1=n+1;
	     int p2=n+1;
	     for(int j=0;j<n;j++)
	      {
	          if(a[j]==1)
	          {
	              p1=j;
	              break;
	          }
	      }
	      for(int j=0;j<n;j++)
	      {
	          if(a[j]==-1)
	          {
	              p2=j;
	              break;
	          }
	      }
	      for(int j=0;j<n;j++)
	      {
	          if(a[j]==b[j])
	          {
	            continue;
	          }
	          else
	          {
	              if(b[j]>a[j])
	              {
	                  if(p1<j)
	                  {
	                      continue;
	                  }
	                  else
	                  {
	                     System.out.println("NO"); 
	                     return;
	                  }
	              }
	              else
	              {
	                if(p2<j)
	                  {
	                      continue;
	                  }
	                  else
	                  {
	                     System.out.println("NO"); 
	                     return;
	                  }  
	              }
	          }
	      }
	      System.out.println("YES");
    }
	public static void main (String[] args) throws IOException {
	    BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
	    StringTokenizer st=new StringTokenizer(br.readLine());
	    int q=Integer.parseInt(st.nextToken());
	  for(int z=1;z<=q;z++)
	  {
	      st=new StringTokenizer(br.readLine());
	      int n=Integer.parseInt(st.nextToken());
	      int a[]=new int[n];
	      int b[]=new int[n];
	      st=new StringTokenizer(br.readLine());
	      for(int j=0;j<n;j++)
	      {
	          a[j]=Integer.parseInt(st.nextToken());
	      }
	      st=new StringTokenizer(br.readLine());
	      for(int j=0;j<n;j++)
	      {
	          b[j]=Integer.parseInt(st.nextToken());
	      }
	      fun(a,b,n);
	  }
	   
	        }
	    }
	       
	   
