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.ArrayList;
import java.util.Scanner;

public class Main {
	
	public static void main(String[] args)
	{
		Scanner s=new Scanner(System.in);
	
		int t=s.nextInt();
		for(int ie=0;ie<t;ie++) {
			
		int n=s.nextInt();
		int[] arr=new int[n];
		for(int i=0;i<n;i++) {
			arr[i]=s.nextInt();
		}
		long[] brr=new long[n];
		for(int i=0;i<n;i++) {
			brr[i]=s.nextLong();
		}
		
		
		int ne=-1;
		int p=-1;
		int flag=0;
			for(int i=0;i<n;i++) {
				if(brr[i]-arr[i]>0) {
					if(p==-1) {
					flag=1;
						System.out.println("NO");
						break;
					}else {
						if(arr[i]<0) {
							ne=1;
						}
					}
				}else if(brr[i]-arr[i]<0) {
					if(ne==-1) {
						flag=1;
						System.out.println("NO");
						break;
						
					}else {
						if(arr[i]>0) {
							p=1;
						}
					}
					
					
					
				}else {
					if(arr[i]>0) {
						p=1;
					}else if(arr[i]<0) {
						ne=1;
					}
				}
			}
			
			if(flag==0) {
				System.out.println("YES");
			}
			
			
		}
		
		
		
	}
}