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.InputStreamReader;
import java.io.PrintWriter;
import java.math.*;
import java.math.BigDecimal;
import java.math.BigInteger;
import java.util.*;
import java.util.Arrays;
import java.util.List;
import java.util.StringTokenizer;
import java.util.function.*;
import java.util.stream.*;


public class B {

    private static final FastReader in = new FastReader();
    private static final FastWriter out = new FastWriter();

    public static void main(String[] args) {
        new B().run();
    }

    private void run() {
        var t = in.nextInt();
        while (t-- > 0) {
            if (solve()) {
                out.println("YES");
            } else {
                out.println("NO");
            }
        }

        out.flush();
    }

    private boolean solve() {
        var n = in.nextInt();
        var a = in.nextIntArray(n);
        var b = in.nextIntArray(n);

        var pos = 0;
        var neg = 0;
        for (var i = 0; i < n; i++) {
            if (a[i] < b[i]) {
                if (pos == 0) return false;
            } else if (a[i] > b[i]) {
                if (neg == 0) return false;
            }
            if (a[i] == 1) pos++;
            else if (a[i] == -1) neg++;
        }

        return true;
    }

}


class FastReader {

    private static final BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    private static StringTokenizer in;

    public String next() {
        while (in == null || !in.hasMoreTokens()) {
            try {
                in = new StringTokenizer(br.readLine());
            } catch (IOException e) {
                return null;
            }
        }
        return in.nextToken();
    }

    public BigDecimal nextBigDecimal() {
        return new BigDecimal(next());
    }

    public BigInteger nextBigInteger() {
        return new BigInteger(next());
    }

    public boolean nextBoolean() {
        return Boolean.valueOf(next());
    }

    public byte nextByte() {
        return Byte.valueOf(next());
    }

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

    public double[] nextDoubleArray(int length) {
        var a = new double[length];
        for (var i = 0; i < length; i++) {
            a[i] = nextDouble();
        }
        return a;
    }

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

    public int[] nextIntArray(int length) {
        var a = new int[length];
        for (var i = 0; i < length; i++) {
            a[i] = nextInt();
        }
        return a;
    }

    public long nextLong() {
        return Long.valueOf(next());
    }

    public long[] nextLongArray(int length) {
        var a = new long[length];
        for (var i = 0; i < length; i++) {
            a[i] = nextLong();
        }
        return a;
    }

}


class FastWriter extends PrintWriter {

    public FastWriter() {
        super(System.out);
    }

    public void println(double[] a) {
        for (var i = 0; i < a.length; i++) {
            print(a[i]);
            print(i + 1 < a.length ? ' ' : '\n');
        }
    }

    public void println(int[] a) {
        for (var i = 0; i < a.length; i++) {
            print(a[i]);
            print(i + 1 < a.length ? ' ' : '\n');
        }
    }

    public void println(long[] a) {
        for (var i = 0; i < a.length; i++) {
            print(a[i]);
            print(i + 1 < a.length ? ' ' : '\n');
        }
    }

    public void println(Object... a) {
        for (var i = 0; i < a.length; i++) {
            print(a[i]);
            print(i + 1 < a.length ? ' ' : '\n');
        }
    }

    public <T> void println(List<T> l) {
        println(l.toArray());
    }

    public void debug(String name, Object o) {
        String value = Arrays.deepToString(new Object[] { o });
        value = value.substring(1, value.length() - 1);
        System.err.println(name + " => " + value);
    }

}
