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.
// package com.company;
import javax.swing.*;
import javax.swing.text.html.HTMLDocument;
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.*;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.lang.*;
/**
 * @author_jigar_nainuji
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        Palindrome solver = new Palindrome();
        int t = in.nextInt();
//        int t=1;
        for(int i=0;i<t;i++) {
            solver.solve(i, in, out);
        }
        out.close();
    }

    static class node{
        int ans;
        int parent;
        boolean bulb;
    }


    static class Palindrome {
        static int mod = 1000000007;
        static long dp[] = new long[10000];

        void solve(int testNumber, InputReader in, PrintWriter out) {
            int n = in.nextInt();
            int one=0,mone=0;
            int a[] = new int[n];
            int b[] = new int[n];
            for(int i=0;i<n;i++) {
                a[i] = in.nextInt();
                if(a[i]==-1)
                {
                    mone++;
                }
                else if(a[i]==1)
                {
                    one ++;
                }
            }
            boolean flag=false;
            for(int i=0;i<n;i++)
                b[i]=in.nextInt();
            for(int i=n-1;i>=0;i--)
            {
                if(a[i]==1)
                    one--;
                if(a[i]==-1)
                    mone--;
                if(b[i]>a[i] && one<=0)
                {
                   flag=true;
                   break;
                }
                 if(b[i]<a[i] && mone<=0)
                {
                    flag=true;
                }


            }
            if(flag)
                System.out.println("NO");
            else
                System.out.println("YES");
        }
    }

    static class Maths{
        static int gcd(int a, int b) {
            if (a == 0)
                return b;
            return gcd(b % a, a);
        }

        public  static int lcm(int a, int b)
        {
            return (a*b)/gcd(a, b);
        }

        public static long factorial(int n){
            long fact = 1;
            for(int i=1;i<=n;i++){
                fact *= i;
            }
            return fact;
        }
    }

    static class Characters{
        public static boolean isVovel(char a){
            return a == 'a' || a == 'e' || a == 'i' || a == 'o' || a == 'u';
        }
    }

    static class Binary {
        public static long numberOfBits(long n){
            long count=0;
            while(n>0){
                count++;
                n >>=1;
            }
            return count;
        }

        public static long numberOfSetBits(long n){
            long count = 0;
            int p = 1;
            while (n>0){
                if((n&p) == 1){
                    count++;
                }
                n >>= 1;
            }
            return count;
        }
    }

    static class InputReader {
        BufferedReader reader;
        StringTokenizer tokenizer;

        InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }

        String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

        int nextInt(){
            return Integer.parseInt(next());
        }

        public Long nextLong(){
            return Long.parseLong(next());
        }
    }
}
