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 may2020;

import java.util.*;
import java.lang.*;
import java.math.*;
import java.io.*;

public class C {
//	static int a,b;
	public static void main(String[] args) throws IOException {
		FastReader scn = new FastReader();
		OutputStream out = new BufferedOutputStream(System.out);
		int t = scn.nextInt();
//		@SuppressWarnings("unchecked")
		while(t-->0) {
			int n =scn.nextInt();
			int[] a = new int[n];
			int[] b = new int[n];
			int[] inc = new int[n];
			int[] dec = new int[n];
			for(int i =0;i<n;i++) {
				a[i] = scn.nextInt();
				if(i==0) {
					inc[0]=a[i]==1?1:0;
					dec[0]=a[i]==-1?1:0;
				}
				if(i>0) {
					if(a[i]==1) {
						inc[i]=inc[i-1]+1;
						dec[i]=dec[i-1];
					}else if (a[i]==-1) {
						dec[i] = dec[i-1]+1;
						inc[i]=inc[i-1];
					}else {
						inc[i]=inc[i-1];
						dec[i]=dec[i-1];
					}
				}
			}
//			for(int i =0;i<n;i++) {
//				System.out.print(inc[i]+" ");
//			}
//			System.out.println();
			for(int i=0;i<n;i++) {
				b[i] = scn.nextInt();
			}
			if(a[0]!=b[0]) {
				out.write("NO\n".getBytes());
				continue;
			}
			boolean flag = true;
			for(int i=1;i<n;i++) {
				if(a[i]!=b[i]) {
					if(a[i]>b[i]) {
						if(dec[i-1]<1) {
							flag = false;
							break;
						}
					}else {
						if(inc[i-1]<1) {
							flag=false;
							break;
						}
					}
				}
			}
			if(flag) {
				out.write("YES\n".getBytes());
			}else {
				out.write("NO\n".getBytes());
			}
		}
//		System.out.println(41664916690999888l/8);
		out.close();
	}
	
//	public static long lcm(long a,long b) {
//		return a/gcd(a,b)*b;
//	}
//	public static long gcd(long a,long b) {
//		if(a == 0)
//			return b;
//		return gcd(b%a,a);
//	}
//	public static int[] merge(int[] one,int[] two){
//		int[] res = new int[one.length+two.length];
//		int i = 0;
//		int j = 0;
//		int k = 0;
//		
//		while(i<one.length&&j<two.length){
//			if(one[i]<two[j]){
//				res[k] = one[i];
//				i++;
//				k++;
//			}
//			else{
//				res[k] = two[j];
//				j++;
//				k++;
//			}
//		}
//		
//		if(i==one.length){
//			while(j<two.length){
//				res[k] = two[j];
//				j++;
//				k++;
//			}
//		}
//		else{
//			while(i<one.length){
//				res[k] = one[i];
//				k++;
//				i++;
//			}
//		}
//		return res;
//	}
//	public static int[] mergesort(int[] arr, int l, int r){
//		if(l==r){
//			int[] br = new int[1];
//			br[0] = arr[l];
//			return br;
//		}
//		int mid = (l+r)/2;
//		int[] fh = mergesort(arr,l,mid);
//		int[] sh = mergesort(arr,mid+1,r);
//		
//		return merge(fh,sh);
//	}
//	static int ub(long[] a, long val) {
//		int l=0, r=a.length;
//		
//		while(r>l) {
//			int x = (r+l)>>1;
//			if(val>=a[x]) {
//				l=x+1;
//			}else {
//				r=x;
//			}
//		}
//		return l;
//	}
//	static int lb(int[] a, int val) {
//		int l = 0;
//		int r = a.length;
//		while(r>l) {
//			int x = (l+r)>>1;
//			if(a[x]>=val) {
//				r=x;
//			}else {
//				l=x+1;
//			}
//		}
//		return l;
//	}
	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; 
        } 
    }

}
