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 D627{
	public static void main(String[] args) throws Exception {
		Scanner sc=new Scanner(System.in);
		int t=sc.nextInt();
             while(t-->0){
                 int n=sc.nextInt();
                 int[] a=new int[n];
                 
                 long[] b=new long[n];
                 int check=10000000;
                 int pos1=check;
                 int pos_1=check;
                 int pos0=check;
                 for(int i=0;i<n;i++){
                     a[i]=sc.nextInt();{
                         if(a[i]<0){
                             pos_1=Math.min(pos_1,i);
                         }
                         else if(a[i]>0){
                             pos1=Math.min(pos1,i);
                         }
                          else if(a[i]==0){
                             pos0=Math.min(pos0,i);
                         }
                     }
                     
                 }
                 for(int i=0;i<n;i++){
                     b[i]=sc.nextLong();
                 }
                 
                // System.out.println(pos1+" "+pos_1+" "+pos0);
                 
                 
                 boolean[] vis=new boolean[n];
                  if(pos0!=check) if(b[pos0]==0){
                     vis[pos0]=true;
                 }
                 
                 
                if(pos1!=check) if(b[pos1]==1){
                     vis[pos1]=true;
                     
                     
                     
                 }
                  
                  if(pos_1!=check)if(b[pos_1]==-1){
                     vis[pos_1]=true;
                 }
                 for(int i=pos1+1;i<n;i++){
                     if(b[i]>=1 && vis[i]==false){
                         vis[i]=true;
                     }
                     else if(i>0){if(b[i]==0 && a[i]==-1 && vis[i]==false){
                         vis[i]=true;
                     }
                     }
                    
                 }
                 
                  for(int i=pos_1+1;i<n;i++){
                     if(b[i]<0 && vis[i]==false){
                         vis[i]=true;
                     }
                     
                      else if(i>0){if(b[i]==0 && a[i]==1 && vis[i]==false){
                         vis[i]=true;
                     }
                     }
                    
                 }
                 
                 
                   for(int i=1;i<n;i++){
                     if(vis[i]==false && b[i]==0 && (a[i])+(a[i-1])==0) vis[i]=true;
                   
                    
                 }
                 
                  for(int i=0;i<n;i++){
                   
                   if(b[i]==0 && a[i]==0){
                       vis[i]=true;
                   }
                   
                     }
                     
                      for(int i=0;i<n;i++){
                   
                   if(b[i]==-1 && a[i]==-1){
                       vis[i]=true;
                   }
                   
                     }
                      for(int i=0;i<n;i++){
                   
                   if(b[i]==1 && a[i]==1){
                       vis[i]=true;
                   }
                   
                     }
                 int ans=0;
                 for(int i=0;i<n;i++){
                     if(vis[i]==false) ans=-1;
                     
                    
                 }
                
                 
                 if(ans==0) System.out.println("yes");
                 else System.out.println("no");
             }
             
           
           }
 }