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.InputMismatchException;
import java.io.IOException;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 * @author Pranay2516
*/public class Main {
	public static void main(String[] args) {
		InputStream inputStream = System.in;
		OutputStream outputStream = System.out;
		FastReader in = new FastReader(inputStream);
		PrintWriter out = new PrintWriter(outputStream);
		TaskB solver = new TaskB();
		int t=in.nextInt();
		for(int q=0;q<t;++q)
		solver.solve(q, in, out);
		out.close();
	}
static class TaskB   {
public void solve(int testNumber, FastReader in, PrintWriter out) {
        int n = in.nextInt();
            int a[] = in.readIntArray(n);
            int b[] = in.readIntArray(n);
            boolean pos=false, neg=false;
            for(int i=0;i<n;++i){
                if(a[i]>b[i] && !neg){
                    out.println("NO");
                    return;
                }
                if(b[i]>a[i] && !pos){
                    out.println("NO");
                    return;
                }
                if(a[i]>0) pos=true;
                else if(a[i]<0) neg=true;
            }
            out.println("YES");
    }

}
static class FastReader   {
private InputStream stream;
private byte[] buf = new byte[1024];
private int curChar;
private int numChars;
private FastReader.SpaceCharFilter filter;

public FastReader(InputStream stream) {
        this.stream = stream;
    }

public int read() {

        if (numChars == -1)
            throw new InputMismatchException();

        if (curChar >= numChars) {

            curChar = 0;

            try {
                numChars = stream.read(buf);
            } catch (IOException e) {
                throw new InputMismatchException();
            }

            if (numChars <= 0)
                return -1;
        }

        return buf[curChar++];
    }

public int nextInt() {

        int c = read();

        while (isSpaceChar(c))
            c = read();

        int sgn = 1;

        if (c == '-') {
            sgn = -1;
            c = read();
        }

        int res = 0;

        do {
            if (c < '0' || c > '9')
                throw new InputMismatchException();

            res *= 10;
            res += c - '0';
            c = read();
        }
        while (!isSpaceChar(c));

        return res * sgn;
    }

public boolean isSpaceChar(int c) {

        if (filter != null)
            return filter.isSpaceChar(c);

        return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
    }

public int[] readIntArray(int size) {
        int[] array = new int[size];
        for (int i = 0; i < size; i++) {
            array[i] = nextInt();
        }
        return array;
    }

public interface SpaceCharFilter   {
public boolean isSpaceChar(int ch);

}

}
}

