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.OutputStream;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Deque;
import java.util.HashMap;
import java.util.HashSet;
import java.util.LinkedList;
import java.util.Queue;
import java.util.Random;
import java.util.StringTokenizer;



public class A
{
	static class Pair
	{
	    int a;
	    int b;
	    Pair(int a,int b)
	    {
	        this.a = a;
	        this.b = b;
	    }
	    public int hashCode(){
	        //System.out.println("In hashcode");
	        int hashcode = 0;
	        hashcode = a*20;
	        hashcode += b*20;
	        return hashcode;
	    }
	     
	    public boolean equals(Object obj){
	        //System.out.println("In equals");
	        if (obj instanceof Pair) {
	            Pair pp = (Pair) obj;
	            return (pp.a == this.a && pp.b == this.b);
	        } else {
	            return false;
	        }
	    
	}
	}
    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;
        }
    }

    static int gcd(int a,int b)
	{
		if(a == 0) return b;
		if(b == 0) return a;
		
		return gcd(b,a%b);
	}
    
    
    
    public static void main(String[] args)
    {
        OutputStream outputStream = System.out;
        FastReader sc = new FastReader();
        PrintWriter out = new PrintWriter(outputStream);
        
        int t = sc.nextInt();
        while(t-- > 0)
        {
        	int n = sc.nextInt();
        	
        	int a[] = new int[n];
        	int b[] = new int[n];
        	
        	for(int i = 0; i < n; i++) a[i] = sc.nextInt();
        	for(int i = 0; i < n; i++) b[i] = sc.nextInt();
        	
        	boolean pos = true;
        	boolean plus = false, minus = false;
        	for(int i = 0; i < n; i++)
        	{	
        		
        		if(i == 0 && a[i] != b[i])
        		{
        			pos = false;
        			break;
        		}
        		if(i != 0)
        		{
        			if(a[i] < b[i] && plus == false)
        			{
        				pos = false;
        				break;
        			}
        			if(a[i] > b[i] && minus == false)
        			{
        				pos = false;
        				break;
        			}
        		}
        	       
        		if(a[i] == 1) plus = true;
        		
        		if(a[i] == -1) minus = true;
        	}
        	
        	if(pos) out.println("YES");
        	else out.println("NO");
        }
        
        out.close();
    }

}