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.*;


import java.math.*;

public class Solution2
{
    public static void main(String arg[]) throws IOException
    {
        CustomInputReader sc = new CustomInputReader();
        PrintWriter pw = new PrintWriter(System.out);
        
        int tc = sc.nextInt();
        while(tc-->0)
        {
            int n = sc.nextInt();
            char s1[] = sc.next().toCharArray();
            char s2[] = sc.next().toCharArray();

            int zeros[] = new int[n];
            int ones[] = new int[n];

            for(int i=0;i<n;++i)
            {
                if(s1[i] == '1') ones[i]++;
                else zeros[i]++;
            }

            for(int i=1;i<n;++i)
            {
                zeros[i]+=zeros[i-1];
                ones[i]+=ones[i-1];
            }

            boolean isPossible = true;

            for(int i=n-1;i>=0;--i)
            {
                if(s1[i] != s2[i])
                {
                    if(zeros[i] != ones[i])
                    {
                        isPossible = false;
                        break;
                    }
                    else
                    {
                        while(i>=0 && s1[i] != s2[i])
                        {
                            i--;
                        }

                        if(i>=0)
                        {
                            if(zeros[i] != ones[i])
                            {
                                isPossible = false;
                                break;
                            }
                            else
                            {
                                i++;
                            }
                        }
                    }
                }
            }

            if(isPossible) pw.println("YES");
            else pw.println("NO");
        }

        //flush output
        pw.flush();
        pw.close();
    }

    static class CustomInputReader 
    {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

        public int nextInt() throws IOException
        {
            return Integer.parseInt(br.readLine());
        }

        public int[] nextIntArr(int n) throws IOException
        {
            String in[] = br.readLine().split(" ");
            int arr[] = new int[n];
            for(int i=0;i<n;++i) arr[i]=Integer.parseInt(in[i]);
            return arr;
        }

        public long nextLong() throws IOException
        {
            return Long.parseLong(br.readLine());
        }

        public long[] nextLongArr(int n) throws IOException
        {
            String in[] = br.readLine().split(" ");
            long arr[] = new long[n];
            for(int i=0;i<n;++i) arr[i]=Long.parseLong(in[i]);
            return arr;
        }

        public String next() throws IOException
        {
            return br.readLine();
        }
    }
        
}

    