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.
// Working program with FastReader 
import java.io.BufferedReader; 
import java.io.IOException; 
import java.io.InputStreamReader; 
import java.util.Scanner; 
import java.util.LinkedList; 
import java.util.StringTokenizer; 

public class Main 
{ 
    
    public static void main(String[] args) 
	{ 
		FastReader sc=new FastReader(); 
		int test=sc.nextInt();
		while(test-->0)
		{
           int n=sc.nextInt();
           int arr[]=new int[n];
           int brr[]=new int[n];
           boolean flag=true;
           int f1a=-1,f1b=-1;
           int foa=-1,fob=-1;
           int fna=-1,fnb=-1;
           for(int i=0;i<n;i++){
               arr[i]=sc.nextInt();
               if(f1a <0 && arr[i] == 1) f1a=i;
               if(foa <0 && arr[i] == 0) foa=i;
               if(fna <0 && arr[i] == -1) fna=i;
           }
           for(int i=0;i<n;i++){
               brr[i]=sc.nextInt();
            //   if(f1b <0 && brr[i] == 1) f1b=i;
            //   if(fob <0 && brr[i] == 0) fob=i;
            //   if(fnb <0 && brr[i] == -1) fnb=i;
           }
           if(arr[0] != brr[0]){
               System.out.println("NO");
               continue;
           }
           for(int i=1;i<n;i++){
               int temp=brr[i];
               if(brr[i] > arr[i]){
                   if(f1a >= i || f1a == -1) {
                       flag=false;
                       break;
                   }
               }
               if(brr[i] < arr[i]){
                   if(fna >= i || fna == -1) {
                       flag=false;
                       break;
                   }
               }
           }
           if(flag){
               System.out.println("YES");
           }
           else  System.out.println("NO");
            
		}
		
	
	} 
	
	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; 
		} 
	} 


} 
