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;
import java.util.Scanner;
import java.util.StringTokenizer;
import java.util.*;




public class CodeForces {

    static class FastReader {
        BufferedReader br;
        StringTokenizer st;
 
        public FastReader()
        {
            br = new BufferedReader(
                new InputStreamReader(System.in));
        }
 
        String next()
        {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                }
                catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }
 
        int nextInt() { return Integer.parseInt(next()); }
 
        long nextLong() { return Long.parseLong(next()); }
 
        double nextDouble()
        {
            return Double.parseDouble(next());
        }
 
        String nextLine()
        {
            String str = "";
            try {
                str = br.readLine();
            }
            catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }
    
    //===================================================================================================================================================

    public static void main(String[] args) {
        FastReader in = new FastReader();
        int t = in.nextInt();
        while(t-->0) {
            int n = in.nextInt();
            String a = in.nextLine();
            String b = in.nextLine();
            a = a + '0';
            b = b + '0';
            int c0 = 0;
            int c1 = 0;
            int f = 1;
            for(int i=0;i<n;i++) {
                if(a.charAt(i) == '0') {
                    c0++;
                } else {
                    c1++;
                }
                
                if((((a.charAt(i) != b.charAt(i)) && (a.charAt(i+1) == b.charAt(i+1))) || ((a.charAt(i) == b.charAt(i)) && a.charAt(i+1) != b.charAt(i+1))) && c0!= c1) {
                    f = 0;
                    System.out.println("NO");
                    break;
                }
            }
            if(f == 1) {
                System.out.println("YES");
            }
        }
    }
}
