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.io.*;
import java.util.*;
public class KindAnton632B {
	  static class FastReader {

	        BufferedReader br;
	        StringTokenizer st;

	        private FastReader() {
	            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());
	        }

		 int[] nextIntArray(int n) {
				int[] a = new int[n];
				for (int i = 0; i < n; i++)
					a[i] = nextInt();
				return a;
		}

	        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;
	        }
	    }
	public static void main(String[] args) {
		FastReader s =new FastReader();
		int t = s.nextInt();
		StringBuilder str = new StringBuilder();
		while(t-- >0) {
			int n = s.nextInt();
			int three[] = s.nextIntArray(n);
			int arr[] = s.nextIntArray(n);
			if(arr[0]!=three[0]) {
				str.append("NO\n");
				continue;
			}
			Set<Integer> set = new HashSet<>();
			
			boolean flag = true;
			for(int i=0;i<n;i++) {
				if(arr[i]>three[i]&&!set.contains(1)) {
					flag = false;
					break;
				}else if(arr[i]<three[i]&&!set.contains(-1)) {
					flag = false;
					break;
				}
				set.add(three[i]);
					
			}
			if(flag)str.append("YES\n");else str.append("NO\n");
		}
		System.out.println(str);

	}

}
