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 Main {
    // For fast input output
    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

        public FastReader() {
            try {
                br = new BufferedReader(new FileReader("input.txt"));
                PrintStream out = new PrintStream(new FileOutputStream("output.txt"));
                System.setErr(new PrintStream(new FileOutputStream("error.txt")));
                System.setOut(out);
            } catch (Exception e) {
                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;
        }
    }

    // end of fast i/o code
    static StringBuilder str = new StringBuilder("");

    public static void main(String[] args) {
        FastReader reader = new FastReader();
        int t = reader.nextInt();
        while (t-- > 0) {
            int n = reader.nextInt();
            int[] a = new int[n];
            int[] b = new int[n];
            for (int i = 0; i < n; i++)
                a[i] = reader.nextInt();
            for (int i = 0; i < n; i++)
                b[i] = reader.nextInt();
            // if(n==1)
            // {
            //     if(a[0]==b[0])
            //         str.append("YES\n");
            //     else
            //         str.append("NO\n");
            // }
            int firstone, firstneg;
            firstone = firstneg = Integer.MAX_VALUE;
            for(int i = 0; i < n; i++)
            {
                if(a[i]==-1)
                {
                    firstneg = i;
                    break;
                }
            }
            for(int i = 0; i < n; i++)
            {
                if(a[i]==1)
                {
                    firstone = i;
                    break;
                }
            }
            int diff = 0;
            boolean illegal = false;
            System.err.println("firstone: "+firstone);
            System.err.println("firstneg: "+firstneg);
            for(int i = 0; i < n; i++)
            {
                diff = b[i] - a[i];
                if(diff>0)
                {
                    if(firstone>=i)
                    {
                        illegal = true;
                        break;
                    }
                }
                if(diff<0)
                {
                    if(firstneg>=i)
                    {
                        illegal = true;
                        break;
                    }
                }
            }
            if(illegal)
                str.append("NO\n");
            else
                str.append("YES\n");            
        }
        System.out.println(str.toString());
    }
}