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.*;
public class Main
{
	public static void main (String[] args) throws java.lang.Exception
	{
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		BufferedWriter bw=new BufferedWriter(new OutputStreamWriter(System.out));
		int T=Integer.parseInt(br.readLine());
		while(T-->0){
		    int n=Integer.parseInt(br.readLine());
		    String a=br.readLine();
		    String b=br.readLine();
		    int oa=0,za=0,ob=0,zb=0;
		    for(int i=0;i<n;i++){
		        if(a.charAt(i)=='0'){
		            za++;
		        }else{
		            oa++;
		        }
		        if(b.charAt(i)=='0'){
		            zb++;
		        }else{
		            ob++;
		        }
		    }
		    int i=n-1;
		    while(i>=0&&a.charAt(i)==b.charAt(i)){
		       if(a.charAt(i)=='0'){
		          za--;
		          zb--;
		       }else{
		          oa--;
		          ob--;
		       }
		       i--;
		    }
		    if(za!=zb||oa!=ob){
		       bw.write("NO\n");
		       continue;
		    }
		    boolean f=true;
		    while(i>=0){
		       if(a.charAt(i)==b.charAt(i)&&oa==ob&&za==zb&&za==oa){
		          while(i>=0&&a.charAt(i)==b.charAt(i)){
		             if(a.charAt(i)=='0'){
		               za--;
		               zb--;
		             }else{
		               oa--;
		               ob--;
		            }
		            i--;
		          }
		       }
		       else if(a.charAt(i)!=b.charAt(i)&&oa==ob&&za==zb&&za==oa){
		          while(i>=0&&a.charAt(i)!=b.charAt(i)){
		             if(a.charAt(i)=='0'){
		               za--;
		               ob--;
		             }else{
		               oa--;
		               zb--;
		            }
		            i--;
		          }
		       }else{
		          f=false;
		          break;
		       }
		    }
		    if(f){
		       bw.write("YES\n");
		    }else{
		       bw.write("NO\n");
		    }
		}
		bw.flush();
	}
}