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.
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define double  long  double
#define F first
#define S second
#define Tsetso ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
const ll N = 1e5+10 , NN= 1e6+10 , length = 25 , mod = 1e9+7;
int n;
string str1 , str2 ;
bool check( )
{
    int ones = 0 , zeros = 0 ;
    for ( int i = 0 ; i < n ; i++)
    {
        ones += str1[i] == '1';
        zeros += str1[i] == '0';
    }
    int last_shift = 0 ;
    while (str1.size() && str2.size())
    {
        if (last_shift) {
            if ( str1.back() == '0')
                    str1.back() = '1';
            else
                    str1.back() = '0';
        }
        if ( str1.back() == str2.back())
        {
            if ( str1.back() == '1')
                ones--;
            else
                zeros--;
        }
        else
        {
            if ( ones == zeros)
            {
                if ( str1.back() == '1')
                    zeros--;
                else
                    ones--;
            }
            else
                return false;
            last_shift++;
        }
        str1.pop_back();
        str2.pop_back();
        last_shift %= 2 ;
    }
    return true ;
}
int main() {
    Tsetso
    int t;
    cin >> t ;
    while(t--)
    {
        cin >> n ;
        cin >> str1 >> str2 ;
        cout << (check() ? "YES\n" : "NO\n");
    }
}