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 QuestionE {

	public static void main(String[] args) throws IOException {
		Reader.init(System.in);
		int t = Reader.nextInt();
		for (int i = 0;i<t;i++) {
			int r=Reader.nextInt();boolean rp = false;boolean chotu = false;int[] arr = new int[r];int[] arr2 = new int[r];
			boolean[] k = new boolean[r];
			boolean[] xy = new boolean[r];
			for (int ij = 0;ij<r;ij++) {
				arr[ij]=Reader.nextInt();
				if (!rp && arr[ij]==1) {
					for (int py = ij+1;py<r;py++) {
						k[py]=true;
					}
					rp=true;
				}
				if (!chotu && arr[ij]==-1) {
					for (int py = ij+1;py<r;py++) {
						xy[py]=true;
					}
					chotu=true;
				}
				
				
			}
			for (int iz = 0;iz<r;iz++) {
				arr2[iz]=Reader.nextInt();
			}
			for (int ab = 0;ab<r;ab++) {
				if (arr2[ab]>arr[ab]) {
					
					if (!k[ab]) {
						System.out.println("NO");break;
					}
				}
				else if (arr2[ab]<arr[ab]) {
					if (!xy[ab]) {
					
						System.out.println("NO");break;
						
					}
				}
				if (ab==(r-1)) {
					System.out.println("YES");
					break;
				}
					
				
			}
			
		}
			
			
			
			
			
		
	
	
	}
	}


class Reader {
    static BufferedReader reader;
    static StringTokenizer tokenizer;
    static void init(InputStream input) {
        reader = new BufferedReader(
                     new InputStreamReader(input) );
        tokenizer = new StringTokenizer("");
    }
    static String next() throws IOException {
        while ( ! tokenizer.hasMoreTokens() ) {
            //TODO add check for eof if necessary
            tokenizer = new StringTokenizer(
                   reader.readLine() );
        }
        return tokenizer.nextToken();
    }

    static int nextInt() throws IOException {
        return Integer.parseInt( next() );
    }
	
    static double nextDouble() throws IOException {
        return Double.parseDouble( next() );
    }
    static long nextLong() throws IOException {
        return Long.parseLong( next() );
    }
}