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.*;
import java.util.StringTokenizer;

public class codeforces712_B {
    private static void solve(FastIOAdapter ioAdapter) {
        int n = ioAdapter.nextInt();
        char[] a = ioAdapter.next().toCharArray();
        char[] b = ioAdapter.next().toCharArray();

        boolean[] good = new boolean[n];
        int count0 = 0;
        int count1 = 0;
        for (int i = 0; i < n; i++) {
            if (b[i] == '0') {
                count0++;
            } else {
                count1++;
            }
            if (count0 == count1) {
                good[i] = true;
            }
        }

        boolean toEqual = true;
        boolean goodAll = true;
        for (int i = n - 1; i >= 0; i--) {
           if (toEqual) {
               if (b[i] != a[i]) {
                   if (!good[i]) {
                       goodAll = false;
                       break;
                   } else {
                       toEqual = false;
                   }
               }
           } else {
               if (b[i] == a[i]) {
                   if (!good[i]) {
                       goodAll = false;
                       break;
                   } else {
                       toEqual = true;
                   }
               }
           }
        }

        if (goodAll) {
            ioAdapter.out.println("YES");
        } else {
            ioAdapter.out.println("NO");
        }
    }

    public static void main(String[] args) throws Exception {
        try (FastIOAdapter ioAdapter = new FastIOAdapter()) {
            int count = 1;
            count = ioAdapter.nextInt();
            while (count-- > 0) {
                solve(ioAdapter);
            }
        }
    }

    static class FastIOAdapter implements AutoCloseable {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        public PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter((System.out))));
        StringTokenizer st = new StringTokenizer("");

        String next() {
            while (!st.hasMoreTokens())
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        int[] readArray(int n) {
            int[] a = new int[n];
            for (int i = 0; i < n; i++) a[i] = nextInt();
            return a;
        }

        long nextLong() {
            return Long.parseLong(next());
        }

        @Override
        public void close() throws Exception {
            out.flush();
            out.close();
            br.close();
        }
    }
}