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.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.util.ArrayList;
import java.util.List;
import java.util.StringTokenizer;

/**
 * Main
 */
public class Main {
    public static void main(String[] args) {
        InputReader ir = new InputReader();
        OutputWriter ow = new OutputWriter();

        solve(ir, ow);

        ow.flush();
    }

    private static void solve(InputReader ir, OutputWriter ow) {
        int t = ir.nextInt();

        while (t-- > 0) {
            int len = ir.nextInt();
            int[] a = new int[len];

            int canAdd = len, canSubtract = len;

            for (int i = 0; i < len; i++) {
                int x = ir.nextInt();

                if (canAdd != len && canSubtract != len) {
                    continue;
                }

                a[i] = x;

                if (x == 1 && canAdd == len) {
                    canAdd = i;
                }

                if (x == -1 && canSubtract == len) {
                    canSubtract = i;
                }
            }

            boolean possible = true;

            for (int i = 0; i < len; i++) {
                int y = ir.nextInt();

                if (i <= canAdd && a[i] < y) {
                    possible = false;
                }

                if (i <= canSubtract && a[i] > y) {
                    possible = false;
                }
            }

            ow.println((possible) ? "YES" : "NO");

        }
    }

}

////////////////// IO //////////////////////

class InputReader {
    private BufferedReader br;
    private StringTokenizer st;

    InputReader() {
        br = new BufferedReader(new InputStreamReader(System.in));
    }

    String nextLine() {
        String s = "";
        try {
            s = br.readLine();
        } catch (IOException e) {
            e.printStackTrace();
        }
        return s;
    }

    String next() {
        while (st == null || !st.hasMoreTokens()) {
            try {
                st = new StringTokenizer(br.readLine());
            } catch (IOException e) {
                e.printStackTrace();
            }
        }

        return st.nextToken();
    }

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

    long nextLong() {
        return Long.parseLong(next());
    }
}

class OutputWriter {
    private BufferedWriter bw;

    OutputWriter() {
        bw = new BufferedWriter(new OutputStreamWriter(System.out));
    }

    void print(String s) {
        try {
            bw.write(s, 0, s.length());
        } catch (IOException e) {
            e.printStackTrace();
        }
    }

    void println() {
        try {
            bw.newLine();
        } catch (IOException e) {
            e.printStackTrace();
        }
    }

    void println(String s) {
        print(s);
        println();
    }

    void print(int x) {
        print(Integer.toString(x));
    }

    void println(int x) {
        println(Integer.toString(x));
    }

    void print(long l) {
        print(Long.toString(l));
    }

    void println(long l) {
        println(Long.toString(l));
    }

    void print(Object o) {
        print(o.toString());
    }

    void println(Object o) {
        println(o.toString());
    }

    void flush() {
        try {
            bw.flush();
        } catch (IOException e) {
            e.printStackTrace();
        }
    }
}
