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.*;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while (t-- > 0) {
            int n = sc.nextInt();
            String a = sc.next(); String b = sc.next();
            int count0 = 0; int count1 = 0;
            if(a.charAt(0)=='1') count1++;
            else count0++;
            boolean isRev = (a.charAt(0)==b.charAt(0));
            boolean isValid = true;
            int sameCount = isRev ? 1 : 0;
            int diffCount  = isRev ? 0 : 1;
            for(int i=1;i<n;i++){
                boolean isEqual = a.charAt(i)==b.charAt(i);
                if(isRev!=isEqual && count0!=count1){
                    isValid = false; break;
                }

                if(count0==count1){
                    sameCount = 0; diffCount = 0;
                }

                if(isEqual) sameCount++;
                else diffCount++;

                if(a.charAt(i)=='1') count1++;
                else count0++;

                isRev = isEqual;
            }
//            System.out.println(diffCount);
            if(diffCount>0 && count0 != count1) isValid = false;
            if(isValid) System.out.println("YES");
            else System.out.println("NO");
        }
    }
}
