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.ArrayList;
import java.util.Collections;
import java.util.HashMap;
import java.util.PriorityQueue;
import java.util.StringTokenizer;
import java.io.*;
 
public class FlipBits {
	
	private static class MyScanner {
	    BufferedReader br;
	    StringTokenizer st;

	    public MyScanner() {
	       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());
	    }

	    String nextLine(){
	        String str = "";
		  try {
		     str = br.readLine();
		  } catch (IOException e) {
		     e.printStackTrace();
		  }
		  return str;
	    }

	 }
	

	
	public static String solution(int n, String a, String b)
			
	{
	   ArrayList<Integer> ind = new ArrayList<>();
	   int one = 0;
	   int zero = 0;
	   
	   for(int i = 0; i<n; i++)
	   {
		   if(a.charAt(i)=='0')
			   zero++;
		   
		   if(a.charAt(i)=='1')
			   one++;
		   
		   if(zero==one)
			   ind.add(i);
			   
	   }
	   
	   int j = 0;
	   
	   for(int i = 0; i<ind.size(); i++)
	   {
		   int curr = ind.get(i);
		   int f = 0;
		   
           if(a.charAt(j)==b.charAt(j))
        	   f=1;
        	   
		   while(j<=curr)
		   {
			   if(f==1)
			   {
				   if(a.charAt(j)!=b.charAt(j))
				     return "NO";
			   }
			   
			   if(f==0)
			   {
				   if(a.charAt(j)==b.charAt(j))
				     return "NO";
			   }
			   
			   j++;
		   }
	   }
	   
	   if(j<n)
	   {
		   for(;j<n; j++)
		   {
			   if(a.charAt(j)!=b.charAt(j))
				     return "NO";
			   
		   }
	   }
	   
	   return "YES";
	}
        
private static PrintWriter out = new PrintWriter(System.out);

public static void main (String[] args)
{
	MyScanner s =  new MyScanner();
     
    int t = s.nextInt();
    
    for(int j = 0; j<t ; j++)
    {
    	int n = s.nextInt();
    	String a = s.nextLine();
    	String b = s.nextLine();
    	
    	out.println(solution(n,a,b));
    }
    
    out.flush();
    out.close();
    
}
}