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<cstdio>
#include<algorithm>
#include<vector>
using namespace std;
char p[6010], q[6010];
int n;
int main() {
	int i, j;
	vector<int>P, Q;
	scanf("%d", &n);
	scanf("%s%s", p,q);
	char cur = '0';
	for (i = 0; p[i]; i++) {
		if (cur != p[i])P.push_back(i);
		cur = p[i];
	}
	if (cur != '0')P.push_back(i);
	cur = '0';
	for (i = 0; q[i]; i++) {
		if (cur != q[i])Q.push_back(i);
		cur = q[i];
	}
	if (cur != '0')Q.push_back(i);
	int Psz = P.size(), Qsz = Q.size();
	int res = 1e9;
	int sss = (n + 1) / 2 * 2;
	for (i = -sss; i <= sss; i+=2) {
		int s = 0;
		for (j = -5000; j <= 15000; j++) {
			int x, y;
			if (i + j < 0)x = 0;
			else if(i + j < Psz)x = P[i + j];
			else x = n;
			if (j < 0)y = 0;
			else if (j < Qsz)y = Q[j];
			else y = n;
			s += abs(x - y);
		}
		res = min(res, s);
	}
	printf("%d\n", res);
}