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.
// cf B
import java.util.*;
// import java.lang.*;
import java.io.*;

//           THIS TEMPLATE MADE BY AKSH BANSAL.

public class Solution {
    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;
        }
    }
    private static boolean[] isPrime;
    private static void primes(){
        int num = (int)1e6; // PRIMES FROM 1 TO NUM
        isPrime = new boolean[num];
     
        for (int i = 2; i< isPrime.length; i++) {
           isPrime[i] = true;
        }
        for (int i = 2; i< Math.sqrt(num); i++) {
           if(isPrime[i] == true) {
              for(int j = (i*i); j<num; j = j+i) {
                 isPrime[j] = false;
              }
           }
        }
    }
    private static long gcd(long a, long b){
        if(b==0)return a;
        return gcd(b,a%b);
    }
    public static void main(String[] args) throws IOException {
        FastReader sc = new FastReader();
        PrintWriter out = new PrintWriter(System.out);
        // primes();
        // ________________________________

        int t = sc.nextInt();
        StringBuilder output = new StringBuilder();

        while (t-- > 0) {
            int n =sc.nextInt();
            String s =sc.nextLine();
            String str =sc.nextLine();
            output.append(solver(s, str, n)).append("\n");
        }

        out.println(output);
        // _______________________________

        // int n = sc.nextInt();
        // out.println(solver());
        // ________________________________
        out.flush();
    }

    public static String solver(String s, String str, int n ) {
        boolean same = true, opp = true;
        int one = 0, zero = 0;
        int len = s.length();
        for(int i=0;i<len;i++){
            char a = s.charAt(i);
            char b = str.charAt(i);

            if(a=='0'){
                zero++;
            }
            else{
                one++;
            }

            if(a==b){
                opp = false;
            }
            else{
                same = false;
            }

            if(one==zero){
                if(opp || same){
                    opp = true;
                    same=  true;
                }
            }
        }
        if(same){
            return "YES";
        }
        else{
            return "NO";
        }
    }
}