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.*;
import java.lang.*;
import java.math.*;
 
public class B {
	public void run() throws Exception {
		FastScanner sc = new FastScanner();
		
		int test = sc.nextInt();
		for (int q = 0; q<test; q++) {
			int n = sc.nextInt();
			long[] arr = new long[n];
			for (int i = 0; i<n; i++) {
				arr[i] = sc.nextLong();
			}
			long[] brr = new long[n];
			for (int i = 0; i<n; i++) {
				brr[i] = sc.nextLong();
			}
			boolean pos = false, neg = false;
			boolean ok = true;
			for (int i =0 ; i<n; i++) {
				if (arr[i] == brr[i]) {
					if (arr[i] == 1) pos = true;
					else if (arr[i] == -1) neg = true;
				}
				else if (arr[i] < brr[i]  && !pos) {
					ok = false;
					break;
				}
				else if (arr[i] > brr[i] && !neg) {
					ok = false;
					break;
				}
				else if (arr[i] == -1) {
					neg = true;
				}
				else if (arr[i] == 1) {
					pos = true;
				}
			}
			if (ok) System.out.println("YES");
			else System.out.println("NO");
		}
		
	}
	static class FastScanner {
		public BufferedReader reader;
		public StringTokenizer tokenizer;
 
		public FastScanner() {
			reader = new BufferedReader(new InputStreamReader(System.in), 32768);
			tokenizer = null;
		}
 
		public String next() {
			while (tokenizer == null || !tokenizer.hasMoreTokens()) {
				try {
					tokenizer = new StringTokenizer(reader.readLine());
				} catch (IOException e) {
					throw new RuntimeException(e);
				}
			}
			return tokenizer.nextToken();
		}
 
		public int nextInt() {
			return Integer.parseInt(next());
		}
 
		public long nextLong() {
			return Long.parseLong(next());
		}
 
		public double nextDouble() {
			return Double.parseDouble(next());
		}
 
		public String nextLine() {
			try {
				return reader.readLine();
			} catch (IOException e) {
				throw new RuntimeException(e);
			}
		}
 
	}
	public static void main (String[] args) throws Exception {
		new B().run();
	}
}