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 Main {

    static Long max = (long) (2 * 10E5);
    private static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    private static BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));

//     Arrays.stream(br.readLine().split("\\s+")).map(Integer::parseInt).toArray(Integer[]::new);
//     Arrays.stream(br.readLine().split("\\s+")).map(Long::parseLong).toArray(Long[]::new);
//     Integer.parseInt(br.readLine());
//     Long.parseLong(br.readLine());
//

    public static void main(String[] args) throws IOException {
        int t = Integer.parseInt(br.readLine());
        for (int _t = 0; _t < t; _t++) {
            int n = Integer.parseInt(br.readLine());
            Integer[] a = Arrays.stream(br.readLine().split("\\s+")).map(Integer::parseInt).toArray(Integer[]::new);
            Integer[] b = Arrays.stream(br.readLine().split("\\s+")).map(Integer::parseInt).toArray(Integer[]::new);
            boolean plus1 = false, minus1 = false;
            String res = "YES";
            for (int i = 0; i < n; i++) {
                if (a[i].equals(b[i])) {
                    if (a[i] == -1)
                        minus1 = true;
                    if (a[i] == 1)
                        plus1 = true;
                } else {
                    if (b[i] > a[i]) {
                        if (!plus1) {
                            res = "NO";
                            break;
                        } else {
                            if (a[i] == -1)
                                minus1 = true;
                        }

                    }
                    if (b[i] < a[i]) {
                        if (!minus1) {
                            res = "NO";
                            break;
                        } else {
                            if (a[i] == 1)
                                plus1 = true;
                        }
                    }
                }
            }
            System.out.println(res);
        }
    }
}