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 MyClass {
    public static void main(String args[]) throws IOException {
      Scanner in = new Scanner(System.in);
      HashSet<Integer> hs = new HashSet<Integer>();
      
      BufferedReader br = new BufferedReader(new InputStreamReader(System.in)); 
      StringTokenizer st;
      
    //   int T = in.nextInt();
      int T = Integer.parseInt(br.readLine());
      
      while(T-- > 0)
      {
         
          hs.clear();
          
          int n = Integer.parseInt(br.readLine());
          
        //   int n = in.nextInt();
          
          int a[] = new int[n];
          int b[] = new int[n];
          
          st = new StringTokenizer(br.readLine());
          for(int i=0 ; i<n ; i++)
          {
            //   a[i] = in.nextInt();
            a[i] = Integer.parseInt(st.nextToken());
              
          }
          
          st = new StringTokenizer(br.readLine());
          for(int i=0 ; i<n ; i++)
          {
            //   b[i] = in.nextInt();
            b[i] = Integer.parseInt(st.nextToken());
          }
          
          if(a[0] != b[0])
          {
              System.out.println("NO");
              continue;
          }
          
          int flag = 1;
          int find;
          
          for(int i=1 ; i<n ; i++)
          {
              if(a[i] == b[i]) continue;
              else if(a[i] > b[i]) find = -1;
              else find = 1;
              
              if(hs.contains(find)) continue;
              
              int f = 0;
              for(int j=i-1 ; j>-1 ; j--)
              {
                  if(a[j] == find)
                  {
                      f=1;
                      hs.add(find);
                      break;
                  }
              }
              
              if(f == 0)
              {
                  flag = 0;
                  break;
              }
              
          }
          
          if(flag == 1) System.out.println("YES");
          else System.out.println("NO");
          
      }
    }
}