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.BufferedInputStream;
import java.util.Scanner;

/**
 * @author hellokiss
 * @since 2021/4/3
 */
public class B {
    public static void main(String[] args) {
        solve();
    }
    public static void solve(){
        Scanner scanner = new Scanner(new BufferedInputStream(System.in));
        int t=scanner.nextInt();
        while(t-->0){
            int n=scanner.nextInt();
            String a=scanner.next();
            String b=scanner.next();
            int sum[]=new int[n];
            for(int i=0;i<a.length();i++){
                if(i==0)sum[i]=a.charAt(i)-'0';
                else sum[i]=sum[i-1]+a.charAt(i)-'0';
            }
            int gs=0;
            boolean isOk=true;
            for(int i=n-1;i>=0;i--){
                if(a.charAt(i)!=b.charAt(i)){
                    if(gs%2==0){
                        if(sum[i]*2==i+1){
                            gs++;
                        }
                        else{
                            isOk=false;
                            break;
                        }
                    }
                    else{

                    }
                }
                else{
                    if(gs%2==0){

                    }
                    else{
                        if(sum[i]*2==i+1){
                            gs++;
                        }
                        else{
                            isOk=false;
                            break;
                        }
                    }
                }
            }
            if(isOk==true){
                System.out.println("YES");
            }
            else {
                System.out.println("NO");
            }
        }
    }
}
