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.*;
import java.math.*;
import java.util.*;
 
 
 
 
 
public class B {
 
    static MyScanner in = new MyScanner();
    static   PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
    static long H,W;
    static BitSet bs;
    static int mod = 1000000007;
    public static void main(String args[]) throws IOException {
   int t = in.nextInt();
   while(t-->0){
    int n= in.nextInt();
    int [] a = new int[n];
    int [] b = new int[n];
    for(int i=0;i<n;i++){
        a[i] = in.nextInt();
    }
    for(int i=0;i<n;i++){
        b[i] = in.nextInt();
    }
    boolean found = true;
    if(a[0]!=b[0]){
        out.println("NO");
    }else{
        int negative =0;
        int positve =0;
        boolean event = true;
        for(int i=0;i<n;i++){
            if(a[i]>b[i]&&negative==0){
                event = false;
                break;
            }else if(a[i]<b[i]&&positve==0){
                event = false;
                break;
            }
            if(a[i]==1){
                positve++;
            } if(a[i]==-1){
                negative++;
            }
        }
        out.println(event?"YES":"NO");
    }
}
 
   out.flush();
    }
    
    private static boolean possible(ArrayList<Integer> list, int b) {

        return false;
    }

    static class IP implements Comparable<IP> {
        public int first,second;
        IP(int first, int second){
            this.first = first;
            this.second = second;
        }
        public int compareTo(IP ip){
            if(first==ip.first)
                return second-ip.second;
            return first-ip.first;    
        }
 
    }
        
       static int gcd(int a, int b){
        return b!=0?gcd(b, a%b):a;
    }
 
        static boolean isEven(long a) {
            return (a&1)==0;
        }
    
    public static class MyScanner {
        BufferedReader br;
        StringTokenizer st;
   
        public MyScanner() {
           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;
        }
  
     }
}