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.util.*;

import java.io.*;

public class Main implements Runnable {

  public static void main(String[] args) {
    new Thread(null, new Main(), "whatever", 1 << 26).start();
  }

  private Scanner sc;
  private StringBuilder sb;

  public void run() {
    sc = new Scanner(System.in);
    sb=new StringBuilder();
    int t = sc.nextInt();
    while (t-- > 0) {
      solve();
    }
    System.out.println(sb);
  }

  public void solve() {
      int n=sc.nextInt();
      char[] a=sc.next().toCharArray();
      char[] b=sc.next().toCharArray();
      
      int[] count=new int[2];
      
      boolean[] p=new boolean[n];
      
      for(int i=0;i<n;i++){
          count[(int)(a[i]-'0')]++;
          if(count[0]==count[1]){
              p[i]=true;
          }
      }
      
      boolean flip=false;
      
      for(int i=n-1;i>=0;i--){
          if((!flip&&a[i]!=b[i])||(flip&&a[i]==b[i])){
              if(p[i]){
                  flip=!flip;
              }else{
                  sb.append("NO\n");
                  return;
              }
          }
      }
      
      sb.append("YES\n");
  }

  class FastScanner {
    private BufferedReader reader = null;
    private StringTokenizer tokenizer = null;

    public FastScanner() {
      reader = new BufferedReader(new InputStreamReader(System.in));
      tokenizer = null;
    }

    public String next() {
      if (tokenizer == null || !tokenizer.hasMoreTokens()) {
        try {
          tokenizer = new StringTokenizer(reader.readLine());
        } catch (IOException e) {
          throw new RuntimeException(e);
        }
      }
      return tokenizer.nextToken();
    }

    public String nextLine() {
      if (tokenizer == null || !tokenizer.hasMoreTokens()) {
        try {
          return reader.readLine();
        } catch (IOException e) {
          throw new RuntimeException(e);
        }
      }
      return tokenizer.nextToken("\n");
    }

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

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

    public double nextDouble() {
      return Double.parseDouble(next());
    }

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

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

    public String[] nextStringArray(int n) {
      String[] a = new String[n];
      for (int i = 0; i < n; i++) {
        a[i] = next();
      }
      return a;
    }

    public long[] nextLongArray(int n) {
      long[] a = new long[n];
      for (int i = 0; i < n; i++) {
        a[i] = nextLong();
      }
      return a;
    }

  }

}
