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.lang.*;
import java.io.*;

public class B {
	public static void main (String[] args) throws java.lang.Exception {
		new Solution();
	}
}
class Solution {
	Reader reader;
	public Solution() {
		reader = new Reader();
		reader.ni();
		while (reader.hasNext()) {
			run_case();
		}
	}
	private void run_case() {
		int n = reader.ni();

		int[] arr_a = new int[n];
		int[] arr_b = new int[n];
		for (int i=0; i<n; i++) arr_a[i] = reader.ni();
		for (int i=0; i<n; i++) arr_b[i] = reader.ni();

		boolean[] flagArr_pos = new boolean[n];
		boolean[] flagArr_neg = new boolean[n];
		Arrays.fill(flagArr_pos, false);
		Arrays.fill(flagArr_neg, false);


		boolean flag_pos = false, flag_neg = false;
		for (int i=0; i<n; i++) {
			if (arr_a[i] == 1) flag_pos = true;
			if (arr_a[i] == -1) flag_neg = true;
			flagArr_pos[i] = flag_pos;
			flagArr_neg[i] = flag_neg;
		}

		int[] diff = new int[n];
		for (int i=0; i<n; i++) diff[i] = arr_b[i] - arr_a[i];

		boolean ok = true;
		for (int i=n-1; i>=0; i--) {
			if (diff[i] == 0) continue;

			if (diff[i] != 0 && i == 0) {
				ok = false;
				break;
			}

			if (diff[i] > 0) {
				if (!flagArr_pos[i-1]) {
					ok = false;
					break;
				}
			} else {
				if (!flagArr_neg[i-1]) {
					ok = false;
					break;
				}
			}
		}

		if (ok) {
			System.out.println("YES");
		} else {
			System.out.println("NO");
		}

		return;
	}
}


class Reader {
	BufferedReader br;
	StringTokenizer st;
	public Reader(){
		try {
			br = new BufferedReader(new InputStreamReader(System.in));
			st = new StringTokenizer(br.readLine());
		} catch (Exception e) {
			e.printStackTrace();
		}
	}
	public String next() {
		if (st.hasMoreTokens()) return st.nextToken();
		try {
			st = new StringTokenizer(br.readLine());
		} catch (Exception e) {
			e.printStackTrace();
		}
		return st.nextToken();
	}
	public boolean hasNext() {
		try {
			if (!st.hasMoreTokens() && !br.ready()) return false;
			else return true;
		} catch (Exception e) {
			e.printStackTrace();
		}
		return false;
	}
	public int ni() {return Integer.parseInt(this.next());}
	public String ns() {return this.next();}
}
