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.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;

public class Main {

    public static void main(String[] args) throws Exception {
        long startTime = System.nanoTime();
        int t = in.nextInt();
        while (t-- > 0) {
            int n = in.nextInt();
            int[] a = new int[n];
            int[] b = new int[n];
            int min = a[0];
            int max = a[0];
            for (int i = 0; i < n; i++) {
                a[i] = in.nextInt();
            }
            for (int i = 0; i < n; i++) {
                b[i] = in.nextInt();
            }
            boolean ans = a[0] == b[0];
            for (int i = 1; i < n; i++) {
                min = Math.min(a[i - 1], min);
                max = Math.max(a[i - 1], max);
                if (a[i] != b[i]) {
                    if (b[i] > a[i] && max == 1) {
                        continue;
                    }
                    if (b[i] < a[i] && min == -1) {
                        continue;
                    }
                    if (b[i] == a[i]) {
                        continue;
                    }
                    ans = false;
                    break;
                }
            }
            if (ans) {
                out.println("YES");
            } else {
                out.println("NO");
            }
        }

        long endTime = System.nanoTime();
        err.println("Execution Time : +" + (endTime - startTime) / 1000000 + " ms");
        exit(0);
    }

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

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 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 long nextLong() {
            return Long.parseLong(next());
        }

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

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

    static void exit(int a) {
        out.close();
        err.close();
        System.exit(a);
    }

    static InputStream inputStream = System.in;
    static OutputStream outputStream = System.out;
    static OutputStream errStream = System.err;
    static InputReader in = new InputReader(inputStream);
    static PrintWriter out = new PrintWriter(outputStream);
    static PrintWriter err = new PrintWriter(errStream);
    static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

}
