You are given strings s and t, both of length N. s and t consist of `0` and `1`. Additionally, in these strings, the same character never occurs three or more times in a row.

You can modify s by repeatedly performing the following operation:

* Choose an index i (1 \leq i \leq N) freely and invert the i-th character in s (that is, replace `0` with `1`, and `1` with `0`), under the condition that the same character would not occur three or more times in a row in s after the operation.



Your objective is to make s equal to t. Find the minimum number of operations required.

Constraints

* 1 \leq N \leq 5000
* The lengths of s and t are both N.
* s and t consists of `0` and `1`.
* In s and t, the same character never occurs three or more times in a row.

Input

Input is given from Standard Input in the following format:


N
s
t


Output

Find the minimum number of operations required to make s equal to t. It can be proved that the objective is always achievable in a finite number of operations.

Examples

Input

4
0011
0101


Output

4


Input

1
0
0


Output

0


Input

8
00110011
10101010


Output

10
#include<iostream>
using namespace std;
const int N = 5010;
int n;
int a[N] , b[N * 3];
char s[N] , t[N];
inline int Abs(int x) { return x < 0 ? -x : x; }
int calc()
{
	int tota = 0 , totb = 0;
	for(int i = 1 ; i < n ; ++i) if(s[i] == '0' && s[i + 1] == '1') a[++tota] = i;
	for(int i = 1 ; i <= n ; ++i) b[++totb] = 0;
	for(int i = 1 ; i < n ; ++i) if(t[i] == '0' && t[i + 1] == '1') b[++totb] = i;
	for(int i = 1 ; i <= n ; ++i) b[++totb] = n;
	int ans = 1e8;
	for(int i = 1 ; i <= totb - tota + 1 ; ++i)
	{
		int res = 0;
		for(int j = 1 ; j < i ; ++j) res += b[j];
		for(int j = 1 ; j <= tota ; ++j) res += Abs(a[j] - b[j + i - 1]);
		for(int j = tota + i ; j <= totb ; ++j) res += n - b[j];
		ans = min(ans , res); 
	}
	return ans;
}

int main()
{
	cin >> n >> (s + 1) >> (t + 1);
	if(n == 1) { cout << (s[1] != t[1]) << '\n'; return 0; }
	if(n == 2) { cout << ((s[1] != t[1]) + (s[2] != t[2])) << '\n'; return 0; }
	int ans = calc();
	for(int i = 1 ; i <= n ; ++i) // 算完红的算蓝的
		s[i] = '1' + '0' - s[i] , t[i] = '1' + '0' - t[i];
	cout << ans + calc() << '\n';
	return 0;
}
/*
2
00
11
*/