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 codechef; // don't place package name! */

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

/* Name of the class has to be "Main" only if the class is public. */
public class Solution
{
	public static void main (String[] args) throws java.lang.Exception
	{
		Scanner sc = new Scanner(System.in);
		
		int t = sc.nextInt();
		l:
		while(t-- > 0) {
		    int n = sc.nextInt();
		    String s = sc.next();
		    String p = sc.next();
		    
		    int [] z = new int[n];
		    int [] o = new int[n];
		    
		    for(int i = 0; i < n; i++) {
		        if(s.charAt(i) == '0') {
		            z[i]++;
		        } else {
		            o[i]++;
		        }
		        
		        if(i > 0) {
		            z[i] += z[i-1];
		            o[i] += o[i-1];
		        }
		    }
		    
		    StringBuilder sb = new StringBuilder();
		    for(int i = 0; i < n; i++) {
		        if(s.charAt(i) == p.charAt(i)) {
		            sb.append('0');
		        } else {
		            sb.append('1');
		        }
		    }
		  //  System.out.println(sb.toString());
		  //  int start = -1;
		  //  int end = -1;
		  //  int count = 0;
		  //  if(s.equals(p)) {
		  //      System.out.println("YES");
		  //  }
		  
		  if(s.equals(p)) {
		      System.out.println("YES");
		      continue l;
		  }
		  
		  //System.out.println(Arrays.toString(o));
		  //System.out.println(Arrays.toString(z));
		    for(int i = 0; i < n; i++) {
		        
		        while(i < n && sb.charAt(i) == '0') {
		            i++;
		        }
		        
		        if(i > 0) {
		            if(o[i-1] != z[i-1] && !(i == n)) {
		                System.out.println("NO");
		                continue l;
		            }
		        }
		        
		        while(i < n && sb.charAt(i) == '1') {
		            i++;
		        }
		        
		        if(i > 0)
		        if(o[i-1] != z[i-1] && !(i == n && sb.charAt(i-1) == '0')) {
	                System.out.println("NO");
	                continue l;
	            }
		    }
		    System.out.println("YES");
		    
		    
		    
		  //  System.out.println(ans);
		}
	}
	
	static boolean isPalindrome(String str)
    {
 
        // Pointers pointing to the beginning
        // and the end of the string
        int i = 0, j = str.length() - 1;
 
        // While there are characters to compare
        while (i < j) {
 
            // If there is a mismatch
            if (str.charAt(i) != str.charAt(j))
                return false;
 
            // Increment first pointer and
            // decrement the other
            i++;
            j--;
        }
 
        // Given string is a palindrome
        return true;
    }
}
