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.BufferedReader;
import java.io.File;
import java.io.FileReader;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

public class b_632 {

	public static void main(String[] args) throws Exception {
		// TODO Auto-generated method stub
		
		FastScanner in = new FastScanner(System.in);
		int T = in.nextInt();
		
		for(int aa = 0; aa < T; aa++) {
			int N = in.nextInt();
			boolean up [] = new boolean [N+1];
			boolean down [] = new boolean [N+1];
			
			int a [] = new int [N];
			int b [] = new int [N];
			
			for(int i = 0; i < N; i++) {
				a[i] = in.nextInt();
				up[i+1] = up[i];
				down[i+1] = down[i];
				if(a[i] ==  1) {
					up[i+1] = true;
				}
				if(a[i] == -1) {
					down[i+1] = true;
				}
			}
			
			for(int i = 0; i < N; i++) {
				b[i] = in.nextInt();
			}
			
			boolean work = true;
			
			for(int i = 0; i < N; i++) {
				int dif = b[i] - a[i];
				if(dif > 0) {
					if(!up[i]) {
						work = false;
					}
				}else if(dif < 0) {
					if(!down[i]) {
						work = false;
					}
				}
			}
			
			if(work) {
				System.out.println("YES");
			}else {
				System.out.println("NO");
			}
		}

	}
	
	static class FastScanner {
		BufferedReader br;
		StringTokenizer st;

		public FastScanner(InputStream stream) {
			br = new BufferedReader(new InputStreamReader(stream));
			st = new StringTokenizer("");
		}

		public FastScanner(String fileName) throws Exception {
			br = new BufferedReader(new FileReader(new File(fileName)));
			st = new StringTokenizer("");
		}

		public String next() throws Exception {
			while (!st.hasMoreTokens()) {
				st = new StringTokenizer(br.readLine());
			}
			return st.nextToken();
		}

		public int nextInt() throws Exception {
			return Integer.parseInt(next());
		}

		public long nextLong() throws Exception {
			return Long.parseLong(next());
		}

		public Double nextDouble() throws Exception {
			return Double.parseDouble(next());
		}

		public String nextLine() throws Exception {
			if (st.hasMoreTokens()) {
				StringBuilder str = new StringBuilder();
				boolean first = true;
				while (st.hasMoreTokens()) {
					if (first) {
						first = false;
					} else {
						str.append(" ");
					}
					str.append(st.nextToken());
				}
				return str.toString();
			} else {
				return br.readLine();
			}
		}
	}

}
