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.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
import java.lang.*;

/* ******************** _/\_Siya pati Ram Chandra ki Jai_/\_ ****************** */

public class JaiShreeRam{
	static Scanner in=new Scanner();
	static int mod = 1000000007;
	public static void main(String[] args) throws Exception{
		int t=in.readInt();
		while(t-->0) {
			int n=in.readInt();
			int a[]=nextIntArray(n);
			int b[]=nextIntArray(n);
			int a1=0,a2=0;
			String s="YES";
			for(int i=0;i<n;i++) {
				if(a[i]>b[i]&&a1==0) {
					s="NO";
					break;
				}
				else if(a[i]<b[i]&&a2==0) {
					s="NO";
					break;
				}
				if(a[i]==-1) {
					a1=1;
				}
				if(a[i]==1)
					a2=1;
			}
			System.out.println(s);
			
		}
	}
	static boolean substring(String s1,int n1, String s2,int n2) {
		for(int i=0;i<=n2-n1;i++) {
			if(s2.substring(i,i+n1).contentEquals(s1))
				return true;
		}
		return false;
	}
	static int[] nextIntArray(int n){
		int[] arr= new int[n];
		int i=0;
		while(i<n){
			arr[i++]=in.readInt();
		}
		return arr;
	}
	static int[] nextIntArray1(int n){
		int[] arr= new int[n+1];
		int i=1;
		while(i<=n){
			arr[i++]=in.readInt();
		}
		return arr;
	}
	static long gcd(long a, long b) {
		if (b==0) return a;
		return gcd(b, a%b);
	}
 
	static void ruffleSort(long[] a) {
		int n=a.length;
		Random r=new Random();
		for (int i=0; i<a.length; i++) {
			int oi=r.nextInt(n); long temp=a[i];
			a[i]=a[oi];
			a[oi]=temp;
		}
		Arrays.sort(a);
	}
	
	static class Scanner {
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer("");
		String readString() {
			while (!st.hasMoreTokens())
				try {
					st=new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			return st.nextToken();
		}
		double readDouble() {
			return Double.parseDouble(readString());
		}
		int readInt() {
			return Integer.parseInt(readString());
		}
		long readLong() {
			return Long.parseLong(readString());
		}
	}	
}
