There is a binary string a of length n. In one operation, you can select any prefix of a with an equal number of 0 and 1 symbols. Then all symbols in the prefix are inverted: each 0 becomes 1 and each 1 becomes 0.

For example, suppose a=0111010000. 

  * In the first operation, we can select the prefix of length 8 since it has four 0's and four 1's: [01110100]00→ [10001011]00. 
  * In the second operation, we can select the prefix of length 2 since it has one 0 and one 1: [10]00101100→ [01]00101100. 
  * It is illegal to select the prefix of length 4 for the third operation, because it has three 0's and one 1. 



Can you transform the string a into the string b using some finite number of operations (possibly, none)?

Input

The first line contains a single integer t (1≤ t≤ 10^4) — the number of test cases.

The first line of each test case contains a single integer n (1≤ n≤ 3⋅ 10^5) — the length of the strings a and b.

The following two lines contain strings a and b of length n, consisting of symbols 0 and 1.

The sum of n across all test cases does not exceed 3⋅ 10^5.

Output

For each test case, output "YES" if it is possible to transform a into b, or "NO" if it is impossible. You can print each letter in any case (upper or lower).

Example

Input


5
10
0111010000
0100101100
4
0000
0000
3
001
000
12
010101010101
100110011010
6
000111
110100


Output


YES
YES
NO
YES
NO

Note

The first test case is shown in the statement.

In the second test case, we transform a into b by using zero operations.

In the third test case, there is no legal operation, so it is impossible to transform a into b.

In the fourth test case, here is one such transformation: 

  * Select the length 2 prefix to get 100101010101. 
  * Select the length 12 prefix to get 011010101010. 
  * Select the length 8 prefix to get 100101011010. 
  * Select the length 4 prefix to get 011001011010. 
  * Select the length 6 prefix to get 100110011010. 



In the fifth test case, the only legal operation is to transform a into 111000. From there, the only legal operation is to return to the string we started with, so we cannot transform a into b.
//package codeforces;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.StringTokenizer;

public class FlipBits {

	static class FastReader{
		BufferedReader br;
		StringTokenizer st;
		FastReader(){
			br=new BufferedReader(new InputStreamReader(System.in));
		}
		String next() throws IOException {
			while(st==null || !st.hasMoreElements()) {
				st=new StringTokenizer(br.readLine());
			}
			return st.nextToken();
		}
		
		int nextInt() throws NumberFormatException, IOException {
			return Integer.parseInt(next());
		}
	}
	
	public static void main(String[] args) throws NumberFormatException, IOException {
		// TODO Auto-generated method stub
		FastReader sc=new FastReader();
		int t=sc.nextInt();
		while(t-->0) {
			int n=sc.nextInt();
			String a=sc.next();
			String b=sc.next();
			int[] legal=new int[n];
			int sum=0;
			for(int i=0;i<n;i++) {
				if(a.charAt(i)=='0') {
					sum+=-1;
				} else {
					sum+=1;
				}
				legal[i]=sum;
			}
			boolean possible=true;
			for(int i=a.length()-1;i>=0;i--) {
				if(i+1<a.length()) {
					if(a.charAt(i)==b.charAt(i) && a.charAt(i+1)!=b.charAt(i+1)) {
						if(legal[i]!=0) {
							possible=false;
							break;
						}
					} else if(a.charAt(i)!=b.charAt(i) && a.charAt(i+1)==b.charAt(i+1)) {
						if(legal[i]!=0) {
							possible=false;
							break;
						}
					}
				} else {
					if(a.charAt(i)==b.charAt(i)) {
						continue;
					} else {
						if(legal[i]!=0) {
							possible=false;
							break;
						}
					}
				}
			}
			if(possible) {
				System.out.println("YES");
			} else {
				System.out.println("NO");
			}
		}
	}

}
