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.util.*;
import java.io.*;

public class B {

    public static void main(String[] args) throws IOException {
        FastReader sc = new FastReader();
        OutputStream outputStream = System.out;
        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];
        	boolean equal = true;
        	boolean one = false, minusOne = false;
        	boolean possible = true;
        	for(int i=0;i<n;i++) a[i] = sc.nextInt();
        	for(int i=0;i<n;i++) {
        		b[i] = sc.nextInt();
        		if(b[i] != a[i]) equal = false;
        	}
        	for(int i=0;i<n;i++) {
        		if(a[i] > b[i] && !minusOne) {
        			possible = false;
        			break;
        		} else if(a[i] < b[i] && !one) {
        			possible = false;
        			break;
        		}
        		if(a[i] == 1) one = true;
        		if(a[i] == -1) minusOne = true;
        	}
        	if(equal || possible) out.println("YES");
        	else out.println("NO");
        }
        out.close();
    }    

    
    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; 
        } 

    }

}