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

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author Nikita Mikhailov
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        FastScanner in = new FastScanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskB2 solver = new TaskB2();
        int testCount = Integer.parseInt(in.next());
        for (int i = 1; i <= testCount; i++)
            solver.solve(i, in, out);
        out.close();
    }

    static class TaskB2 {
        public void solve(int testNumber, FastScanner in, PrintWriter out) {
            int n = in.readInt();
            int a[] = in.readInts(n);
            int[] b = in.readInts(n);

            boolean canAdd = false;
            boolean canRem = false;

            boolean fail = false;
            for (int i = 0; i < n; i++) {
                if (a[i] == b[i]) {
                    // ok
                } else {

                    if (a[i] < b[i]) {
                        if (!canAdd) {
                            fail = true;
                        }
                    }
                    if (a[i] > b[i]) {
                        if (!canRem) {
                            fail = true;
                        }
                    }
                }

                canAdd = canAdd || a[i] == 1;
                canRem = canRem || a[i] == -1;

                if (fail) {
                    break;
                }
            }
            out.println(fail ? "NO" : "YES");
        }

    }

    static class FastScanner {
        private StringTokenizer st;
        private BufferedReader in;

        public FastScanner(final InputStream in) {
            this.in = new BufferedReader(new InputStreamReader(in));
        }

        public String readToken() {
            while (st == null || !st.hasMoreTokens()) {
                try {
                    st = new StringTokenizer(in.readLine());
                } catch (final IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return st.nextToken();
        }

        public int readInt() {
            return Integer.parseInt(readToken());
        }

        public int[] readInts(final int n) {
            final int[] res = new int[n];
            for (int i = 0; i < n; i++) {
                res[i] = readInt();
            }
            return res;
        }

        public String next() {
            return readToken();
        }

    }
}

