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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class Div2 {
  public static void main(String[] args) throws IOException {
    BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
    String[] s = reader.readLine().split(" ");
    StringBuilder sb = new StringBuilder();
    Div2 div2 = new Div2();
    int t = Integer.parseInt(s[0]);
    while (t-- > 0) {
      s = reader.readLine().split(" ");
      int n = Integer.parseInt(s[0]);
      int[] flip = new int[n];
      int[] target = new int[n];
      char[] str = reader.readLine().toCharArray();
      int[] count0 = new int[n];
      int[] count1 = new int[n];
      for (int i = 0; i < n; i++) {
        if (i > 0) {
          count0[i] = count0[i - 1];
          count1[i] = count1[i - 1];
        }
        if ('1' == str[i]) {
          flip[i]++;
          count1[i]++;
        } else {
          count0[i]++;
        }
      }
      str = reader.readLine().toCharArray();
      for (int i = 0; i < n; i++) {
        if ('1' == str[i]) {
          target[i]++;
        }
      }
      boolean valid = true;
      int sum = 0;
      int preEnd = n;
      while (valid) {
        int i = preEnd - 1;
        while (i >= 0 && (flip[i] + sum) % 2 == target[i]) {
          i--;
        }
        if (i < 0) {
          break;
        }
        sum++;
        int tmp0;
        int tmp1;
        if (sum % 2 == 0) {
          tmp0 = count0[i];
          tmp1 = count1[i];
        } else {
          tmp0 = count1[i];
          tmp1 = count0[i];
        }
        if (tmp0 != tmp1) {
          valid = false;
        }
        preEnd = i;
      }
      if (valid) {
        sb.append("YES\n");
      } else {
        sb.append("NO\n");
      }
    }
    System.out.println(sb.toString());
  }
}