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 jan.longB;

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

public class C {


    public static void main(String[] args) {
        try {
            BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(System.in));
            BufferedOutputStream bufferedOutputStream = new BufferedOutputStream(System.out);
            String eol = System.getProperty("line.separator");
            byte[] eolb = eol.getBytes();
            String[] str = bufferedReader.readLine().split(" ");
            int t = Integer.parseInt(str[0]);
            while(t-- > 0) {
                int n = Integer.parseInt(bufferedReader.readLine().split(" ")[0]);
                char[] a = bufferedReader.readLine().split(" ")[0].toCharArray();
                char[] b = bufferedReader.readLine().split(" ")[0].toCharArray();
                int azct, aoct;
                azct = aoct = 0;
                String ans = "YES";
                boolean opp = true, same = true;
                for(int i = 0; i < n; i++) {
                    if(a[i] == '1') aoct++; else azct++;
                    if(a[i] != b[i]) same = false; else opp = false;
                    if(aoct == azct) {
                        if(same || opp) same = opp = true;
                    }
                }
                if(!same) ans = "NO";
                bufferedOutputStream.write(ans.getBytes());
                bufferedOutputStream.write(eolb);
            }
            bufferedOutputStream.flush();
            bufferedOutputStream.close();
        } catch (IOException ioe) {
            ioe.printStackTrace();
        }
    }
}
