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.FileInputStream;
import java.io.FileNotFoundException;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.StringTokenizer;

public class CF1333B {

    public static void main(String[] args) throws Exception {

        InputReader ir = new InputReader();
        int testCases = ir.nextInt();
        for (int testCase = 1; testCase <= testCases; testCase++) {
            solve(ir);

        }

    }

    private static void solve(InputReader ir) {
        int n = ir.nextInt();
        int[] ai = new int[n];
        int[] bi = new int[n];

        for (int i = 0; i < n; i++) {
            ai[i] = ir.nextInt();
        }
        for (int i = 0; i < n; i++) {
            bi[i] = ir.nextInt();
        }
//        System.err.println(Arrays.toString(ai));
//        System.err.println(Arrays.toString(bi));

        boolean neg1Seen = false, pos1Seen = false;

        for (int i = 0; i < n; i++) {
            if (ai[i] < bi[i]) {
                if (pos1Seen == false) {
//                    System.err.println("pos i " + i + " a[i] " + ai[i] + ", b[i] " + bi[i]);
                    System.out.println("NO");
                    return;
                }
            }
            if (ai[i] > bi[i]) {
                if (neg1Seen == false) {
//                    System.err.println("neg i " + i + " a[i] " + ai[i] + ", b[i] " + bi[i]);
                    System.out.println("NO");
                    return;
                }
            }
            if (ai[i] == 1) pos1Seen = true;
            if (ai[i] == -1) neg1Seen = true;
        }
        System.out.println("YES");
        return;
    }

    static class InputReader {
        public BufferedReader reader;
        public StringTokenizer st;

        public InputReader() throws FileNotFoundException {
        reader = new BufferedReader(new InputStreamReader(System.in));
            // reader = new BufferedReader(new InputStreamReader(new FileInputStream("CF1333B.in")));
        }

        public String next() {
            while (st == null || !st.hasMoreTokens()) {
                st = new StringTokenizer(nextLine());
            }
            return st.nextToken();
        }

        public String nextLine() {
            try {
                return reader.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return null;
        }

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