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.util.StringTokenizer;

public class q2 {

	public static void main(String[] args) {
		FastReader s = new FastReader();
		
		int t = s.nextInt();
		
		while(t-- > 0)
		{
			int n = s.nextInt();
			int[] a = new int[n];
			int[] b = new int[n];
			char[] help = new char[n];
			for(int i=0;i<n;++i)
				a[i] = s.nextInt();
			help[0] = 'e';
			for(int i=1;i<n;++i)
			{
				if(help[i-1] == 'e')
				{
					if(a[i-1] == 0) help[i] = 'e';
					else if(a[i-1] == 1) help[i] = 'p';
					else help[i] = 'n';
				}
				else if(help[i-1] == 'p')
				{
					if(a[i-1] == -1) help[i] = 'b';
					else if(a[i-1] == 1) help[i] = 'p';
					else help[i] = help[i-1];
				}
				else if(help[i-1] == 'n')
				{
					if(a[i-1] == 1) help[i] = 'b';
					else if(a[i-1] == -1) help[i] = 'n';
					else help[i] = help[i-1];
				}
				else help[i] = help[i-1];
				
			}
			
			for(int i=0;i<n;++i)
				b[i] = s.nextInt();
			
			int flag = 0;
			if(a[0] != b[0])
			{
				System.out.println("NO");
				continue;
			}
			else
			{
				for(int i=1;i<n;++i)
				{
					if(a[i] < b[i])
					{
						if(help[i] == 'e' || help[i] == 'n')
						{
							flag = 1;break;
						}
					}
					else if(a[i] > b[i])
					{
						if(help[i] == 'e' || help[i] == 'p')
						{
							flag = 1;break;
						}
					}
				}
			}
			
			if(flag == 1)
				System.out.println("NO");
			else 
				System.out.println("YES");
			
			
		}

	}
	static class FastReader {
        BufferedReader br;
        StringTokenizer st;
 
        public FastReader()
        {
            br = new BufferedReader(new
                    InputStreamReader(System.in));
        }
 
        String next()
        {
            while (st == null || !st.hasMoreElements())
            {
                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());
        }
 
        double nextDouble()
        {
            return Double.parseDouble(next());
        }
 
        String nextLine()
        {
            String str = "";
            try
            {
                str = br.readLine();
            }
            catch (IOException e)
            {
                e.printStackTrace();
            }
            return str;
        }
	}
}

