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.lang.*;
public class Codeforces {
    static Scanner sr = new Scanner(System.in);
    static long gcd(long a ,long b) {
        if (b == 0)
            return a;
        return gcd(b, a % b);
    }
    public static void main(String[] args) throws java.lang.Exception {
        StringBuffer ans = new StringBuffer("");
        long T=sr.nextLong();
        while (T-->0) {
            int n=sr.nextInt();
            String s=sr.next();
            String b=sr.next();
            int a[]=new int[n];
            int c=0;
            for(int i=0;i<n;i++)
            {
                if(s.charAt(i)=='1')
                {
                    ++c;
                    a[i]=c;
                }
                else
                    a[i]=c;
            }
            int flag=0;
            boolean ch=false;
            for(int i=n-1;i>-1;i--)
            {
                if(s.charAt(i)!=b.charAt(i))
                {
                    if(ch==false) {
                        if(a[i]!=i+1-a[i]) {
                            flag = -1;
                            break;
                        }
                        else
                        {
                            ch=!ch;
                        }
                    }
                }
                else
                {
                    if(ch==true) {
                        if(a[i]!=i+1-a[i]) {
                            flag = -1;
                            break;
                        }
                        else
                        {
                            ch=!ch;
                        }
                    }
                }
            }
            if(flag==0)
                ans.append("YES"+'\n');
            else
                ans.append("NO"+'\n');
        }
        System.out.println(ans);
    }
}