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.io.*;
import java.util.ArrayList;
import java.util.Arrays;

public class KindAnton {
    public static void main(String[] args) throws IOException {
//                        BufferedReader f = new BufferedReader(new FileReader("C:\\Users\\Owner\\CodeForces\\src\\test.in"));
//        PrintWriter out = new PrintWriter(new BufferedWriter(new FileWriter("C:\\Users\\Owner\\CodeForces\\src\\test.out")));
        BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter out = new PrintWriter(System.out);

        int t = Integer.parseInt(f.readLine());
        for(int d = 0; d < t; d++){
            //test cases
            int n = Integer.parseInt(f.readLine());
            int[] a = new int[n];
            int[] b = new int[n];
            int[] avail = new int[n];
            String[] raw = f.readLine().split(" ");
            for(int i = 0; i < n; i++){
                a[i] =  Integer.parseInt(raw[i]);
            }
            raw = f.readLine().split(" ");
            for(int i = 0; i < n; i++){
                b[i] = Integer.parseInt(raw[i]);
            }

//            System.out.println("a " + Arrays.toString(a));
//            System.out.println("b " + Arrays.toString(b));
//            System.out.println(Arrays.toString(avail));
            boolean done = false;
            if(a[0] != b[0]){
                out.println("NO");
            }
            else{
                avail[0] = a[0];
                for(int i =1; i < n; i++){
                    if(avail[i - 1] == 2){
                        avail[i] = 2;
                    }
                    else if(avail[i - 1] == -1 || avail[i - 1 ] == 1){
                        if(a[i] == -1 * (avail[i - 1])){
                            avail[i] = 2;
                        }
                        else{
                            avail[i] = avail[i - 1];
                        }
                    }
                    else{
                        avail[i] = a[i];
                    }
                }
                for(int i = 1; i < n; i++){
                    if(a[i] - b[i] < 0){
                        if(avail[i - 1] < 1){
                            out.println("NO");
                            done = true;
                            break;
                        }
                    }
                    if(a[i] - b[i] > 0){
                        if(avail[i - 1] != -1 && avail[i - 1] != 2){
                            out.println("NO");
                            done = true;
                            break;
                        }
                    }
                }
                if(!done){
                    out.println("YES");
                }
            }


        }


        out.close();

    }
}
