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.StringTokenizer;

public class FlipsTheBits {
    public static void main(String[] args) {
        FastReader fs = new FastReader();
        StringBuilder sb = new StringBuilder();
        int t = fs.nextInt();
        for(int tt = 0; tt < t; tt++) {
            int n = fs.nextInt();
            String s1 = fs.next();
            String s2 = fs.next();
            boolean flipped = false;
            int ones = 0;
            int zeros = 0;
            for(int i = 0; i < n; i++) {
                if(s1.charAt(i) == '1') ones++;
                else zeros++;
            }
            int counter = 0;
            for(int i = n - 1; i >= 0; i--) {
                char left = s1.charAt(i);
                char right = s2.charAt(i);
                if(flipped) {
                    if(left == '1') left = '0';
                    else left = '1';
                }
                if(left != right && ones != zeros) break;
                else if(left != right) {
                    int temp = ones;
                    ones = zeros;
                    zeros = temp;
                    flipped = !flipped;
                    left = left == '1' ? '0' : '1';
                }
                if(left == '1') ones--;
                else zeros--;
                counter++;
            }
            if(counter == n) sb.append("YES\n");
            else sb.append("NO\n");
        }
        System.out.print(sb);
    }

    public static long div(long a, long b) {
        return (a + b - 1) / b;
    }

    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());
        }

        float nextFloat() {
            return Float.parseFloat(next());
        }
    }

}
