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.*;
public class Main
{
    public static void main(String srgd[])
    {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while(t-- > 0)
        {
            int n = sc.nextInt();
            String a = sc.next();
            String b = sc.next();
            System.out.println(solve(n,a,b));
        }
    }
    
    public static String solve(int n,String a,String b)
    {
        ArrayList<ArrayList<Integer>> list = new ArrayList<>();
        char ar[] = a.toCharArray();
        char br[] = b.toCharArray();
        int i,k=0,one=0,zero=0,j=0;
        for(i=0;i<n;i++)
        {
            if(ar[i]=='0') zero++;
            else
            one++;
            if(one==zero)
            {
                list.add(new ArrayList<>());
                list.get(k).add(j);
                list.get(k).add(i);
                j = i+1;
                k++;
            }
        }
        for(i=0;i<list.size();i++)
        {
            if(ar[list.get(i).get(0)]!=br[list.get(i).get(0)])
            {
                for(int l=list.get(i).get(0);l<=list.get(i).get(1);l++)
                {
                    if(ar[l]=='0') ar[l] = '1';
                    else
                    ar[l]='0';
                }
            }
        }
        for(i=0;i<n;i++)
        {
            if(ar[i]!=br[i])
            return "NO";
        }
        return "YES";
    }
}