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 java.lang.Exception
	{
		MyScanner sc = new MyScanner();
	
		 //  hash=new Hashtable<>();
		 int nn=Integer.parseInt(sc.nextLine());
		 while(nn-->0){
		   int a=Integer.parseInt(sc.nextLine());
		  
		   
		   String hh[]=sc.nextLine().split(" ");
		   String gg[]=sc.nextLine().split(" ");
		   int pos=0;
		   int neg=0;
		   for(int i=0;i<a;i++)
		   {
		      if(Integer.parseInt(hh[i])<0){neg+=1;}
		      else if(Integer.parseInt(hh[i])>0){pos+=1;}
		      
		    //  arr2[i]=Integer.parseInt(gg[i])-Integer.parseInt(hh[i]);
		   }
		   int flag=0;
		 for(int i=a-1;i>=0;i--)
		 {
		    	      if(Integer.parseInt(hh[i])<0){neg-=1;}
		      else if(Integer.parseInt(hh[i])>0){pos-=1;}  
		    if((Integer.parseInt(gg[i])>Integer.parseInt(hh[i]) && pos==0 ) ||(Integer.parseInt(gg[i])<Integer.parseInt(hh[i]) && neg==0)){flag=1;break;}
	
		 }
		 if(flag==1){System.out.println("NO");}
		 else{System.out.println("YES");}
		   
		
	}

	
	}
}
   class MyScanner {
      BufferedReader br;
      StringTokenizer st;
 
      public MyScanner() {
         br = new BufferedReader(new InputStreamReader(System.in));
      }
 
      String next() {
          while (st == null || !st.hasMoreElements()) {
              try {
                  st = new StringTokenizer(br.readLine());
              } catch (IOException e) {
                  e.printStackTrace();
              }
          }
          return st.nextToken();
      }
 
      int nextInt() {
          return Integer.parseInt(next());
      }
 
      long nextLong() {
          return Long.parseLong(next());
      }
 
      double nextDouble() {
          return Double.parseDouble(next());
      }
 
      String nextLine(){
          String str = "";
	  try {
	     str = br.readLine();
	  } catch (IOException e) {
	     e.printStackTrace();
	  }
	  return str;
      }
 
   }