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.io.PrintWriter;
import java.util.Scanner;
import java.util.StringTokenizer;
import java.util.ArrayList;
import static java.lang.Math.min;
import static java.lang.Math.max;
import static java.lang.Math.abs;

// thanks be to Lord and Mary
public class sol {

    static PrintWriter out = new PrintWriter(System.out);
    static FastScanner in = new FastScanner();


    static void solve () {
        int l = in.nextInt();
        String n = in.next();
        String k = in.next();
        boolean ans = true;

        int o = 0;
        int t = 0;
        int flip = 0;
        for(int i = 0;i< l;i++ ) {
            if(n.charAt(i) == '1') o++;
            else t++;

        }
        for(int i = l-1;i>=0;i-- ) {
            char v = n.charAt(i);

            if(flip%2 == 1) {
                if(v == '1') v = '0';
                else v = '1';
            }

            if(v!= k.charAt(i)) {
            if(o==t) {
            flip++;
            } else {
                ans = false;
                break;
            }
            }

            if(n.charAt(i) == '1') o--;
            else t--;

        }



        out.println(ans? "YES":"NO");
    }



    //DO NOT FORGET TO DELETE PACKAGE NAME!!!
    //also don't forget to check the value t
    //use long or int as needed







    public static void main(String[] args ) {
        int t = 1;
        t = in.nextInt();
        //double sTime = System.nanoTime();
        for(int i = 0;i< t;i++ ) {
            solve();
            //double eTime = System.nanoTime();
            //out.println((eTime-sTime)/1000000);
        }

        out.close();
    }








    static class FastScanner {
        BufferedReader br;
        StringTokenizer st;

        public FastScanner() {
            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());
        }
    }
}
// Who am I?
// in nomine patris et filii spiritus sancti. Amen.