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.math.BigInteger;
import java.text.SimpleDateFormat;
import java.time.Instant;
import java.time.LocalDate;
import java.time.LocalDateTime;
import java.time.LocalTime;
import java.util.*;


public class Main {
	 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()); 
	        } 
	  
	        String nextLine() 
	        { 
	            String str = ""; 
	            try
	            { 
	                str = br.readLine(); 
	            } 
	            catch (IOException e) 
	            { 
	                e.printStackTrace(); 
	            } 
	            return str; 
	        } 
	    }
	 
	 static boolean isPalindrome(String str)
	    {
	 
	        // Pointers pointing to the beginning
	        // and the end of the string
	        int i = 0, j = str.length() - 1;
	 
	        // While there are characters to compare
	        while (i < j) {
	 
	            // If there is a mismatch
	            if (str.charAt(i) != str.charAt(j))
	                return false;
	 
	            // Increment first pointer and
	            // decrement the other
	            i++;
	            j--;
	        }
	 
	        // Given string is a palindrome
	        return true;
	    }
	 
	
	public static void main(String[] args) throws IOException {
      FastReader sc=new FastReader();
      int t=sc.nextInt();
      while(t-->0) {
    	  int n=sc.nextInt();
    	  String a=sc.nextLine();
    	  String b=sc.nextLine();
    	  List<List<Integer>> list=new ArrayList<>();
    	  int j=0;
    	  int count1=0,count0=0;
    	  char ch[]=a.toCharArray();
    	  for(int i=0;i<n;i++) {
    		  if(ch[i]=='1') {
    			  count1++;
    		  }
    		  else {
    			  count0++;
    		  }
    		  if(count0==count1) {
    			  List<Integer> temp=new ArrayList<>();
    			  temp.add(j);
    			  temp.add(i);
    			  list.add(temp);
    			  j=i+1;
    		  }
    	  }
    	  
    	  for(int i=0;i<list.size();i++) {
    		  List<Integer> temp=list.get(i);
    		  int start=temp.get(0);
    		  int end=temp.get(1);
    		  if(a.charAt(start)==b.charAt(start)) {
    			  continue;
    		  }
    		  else {
    			  for(int k=start;k<=end;k++) {
    				  if(ch[k]=='0') {
    					  ch[k]='1';
    				  }
    				  else {
    					 ch[k] ='0';
    				  }
    			  }
    		  }
    	  }
    	  
    	  StringBuilder sb=new StringBuilder();
    	  for(int i=0;i<n;i++) {
    		  sb.append(ch[i]);
    	  }
    	  if(sb.toString().equals(b)) {
    		  System.out.println("YES");
    	  }
    	  else {
    		  System.out.println("NO");
    	  }
    	  
      }
	}

}
