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 pb push_back
#define siz(n) n.size()
#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, siz1, siz2, ans;
char s[5010], t[5010];
vector<int> S, T;
int main() {
	cin >> n; scanf("%s%s", s + 1, t + 1);
	for(register int i = 1; i < n; ++i) if(s[i] != s[i + 1]) S.pb(i);
	for(register int i = 1; i < n; ++i) if(t[i] != t[i + 1]) T.pb(i);
	siz1 = siz(S); siz2 = siz(T); ans = 0x3f3f3f3f;
	for(register int p = -siz1 - 1; p <= siz2 + 1; ++p) {
		if((p + (s[1] == t[1])) & 1) {
			int l = 0;
			for(register int i = dmin(-p, 0); i <= dmax(siz1, siz2 - p); ++i) {
				int r = i + p, _p, _c;
				_p = i < 0 ? 0 : (i >= siz1 ? n : S[i]); _c = r < 0 ? 0 : (r >= siz2 ? n : T[r]); l += dabs(_p - _c);
			}
			cmin(ans, l);
		}
	}
	printf("%d\n", ans);
	return 0;
}