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 codeforce.div2.r712;

import java.util.Arrays;
import java.util.Scanner;

/**
 * 5
 * 10
 * 01 00 10 1100
 * 01 00 10 1100
 * 
 * 10 11 0 1 00 00
 * 01 00 1 0 11 00
 * <p>
 * <p>
 * 4
 * 0000
 * 0000
 * 3
 * 001
 * 000
 * <p>
 * 12
 * 1010 10 01 1010
 * 1001 10 01 1010
 * 6
 * 1110 00
 * 1101 00
 *
 * @author pribic (Priyank Doshi)
 * @since 03/04/21
 */
public class B {

  public static void main(String[] args) {
    try (Scanner sc = new Scanner(System.in)) {
      int T = sc.nextInt();
      for (int tt = 1; tt <= T; tt++) {
        int n = sc.nextInt();
        String a = sc.next();
        String b = sc.next();
        int[] diff = new int[n];
        int curDiff = 0;
        for (int i = 0; i < n; i++) {
          if (a.charAt(i) == '1') {
            curDiff++;
          } else {
            curDiff--;
          }
          diff[i] = curDiff;
        }
        //System.out.println("Arrays.toString(diff) = " + Arrays.toString(diff));
        boolean possible = true;
        int flips = 0;
        for (int i = n - 1; i >= 0; i--) {
          if (mapFlip(a.charAt(i), flips) != b.charAt(i)) {
            if (diff[i] != 0) {
              possible = false;
              break;
            } else {
              flips++;
            }
          }

        }
        System.out.println(possible ? "YES" : "NO");
      }
    }
  }

  private static char mapFlip(char c, int flips) {
    if (flips % 2 == 0) return c;
    else return c == '1' ? '0' : '1';
  }
}