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 <bits/stdc++.h>
#define dmin(i, j) ((i) < (j) ? (i) : (j))
#define dmax(i, j) ((i) > (j) ? (i) : (j))
#define cmin(i, j) ((i) > (j) ? (i) = (j) : (i))
#define dabs(i) ((i) < 0 ? -(i) : (i))
using namespace std;
int n, ST[10000010 << 1], *st = ST + 10000010;
long long ans, sum;
char s[10000010], t[10000010];
int main() {
	cin >> n; scanf("%s%s", s, t); ans = 1e18; int _p = 0, _c = (s[0] != t[0]);
	for(register int i = 0; i < n; ++i) {
    	++st[_p - _c]; sum += _p - _c + n + 2;
    	if(s[i] != s[i + 1]) ++_p;
    	if(t[i] != t[i + 1]) ++_c;
    }
    _p = -n;
	for(register int i = -n - 2; i <= n + 2; ++i) {
		if(~i & 1) cmin(ans, sum);
//        cout << sum << endl;
    	_p += st[i] << 1, sum += _p;
    }
	printf("%lld\n", ans);
	return 0;
}