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

public class Sepcode
{
  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 p = sc.nextInt();
        String a = sc.next();
        String b = sc.next();
        int n = p;
        int [] zero = new int[n];
        int [] one = new int[n];
        
        for(int i = 0; i < n; i++) {
            if(a.charAt(i) == '0') {
                zero[i]++;
            } else {
                one[i]++;
            }
            
            if(i > 0) {
                zero[i] += zero[i-1];
                one[i] += one[i-1];
            }
        }
        
        StringBuilder sb = new StringBuilder();
        
        for(int i = 0; i < n; i++) {
            if(a.charAt(i) == b.charAt(i)) {
                sb.append('0');
            } else {
                sb.append('1');
            }
        }
      
      if(a.equals(b)) {
          System.out.println("yes");
          continue l;
      }
        for(int i = 0; i < n; i++) {
            
            while(i < n && sb.charAt(i) == '0') {
                i++;
            }
            
            if(i > 0) {
                if(one[i-1] != zero[i-1] && !(i == n)) {
                    System.out.println("no");
                    continue l;
                }
            }
            
            while(i < n && sb.charAt(i) == '1') {
                i++;
            }
            
            if(i > 0)
            if(one[i-1] != zero[i-1] && !(i == n && sb.charAt(i-1) == '0')) {
                  System.out.println("no");
                  continue l;
              }
        }
        System.out.println("yes");
        
    }
  }
  
  
}