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 <cstring>
#include <algorithm>

using namespace std;

const int N = 2e4 + 7;

int n,a[N], b[N];
char s[N], t[N];

int calc() 
{
	int t1 = 0, t2 = 0;
	for (int i = 1; i < n; ++i)
		if (s[i] == '0' && s[i + 1] =='1') a[++t1] = i;
	for (int i = 1; i <= n; ++i) b[++t2] = 0;
	for (int i = 1; i < n; ++i) 
		if (t[i] == '0' && t[i + 1] == '1') b[++t2] = i;
	for (int i = 1; i <= n; ++i) b[++t2] = n;
	int ret = n * n;
	for (int i = 1; i <= t2 - t1 + 1; ++i)
	{
		int now = 0;
		for (int j = 1; j <= i - 1; ++j) now += b[j];
		for (int j = 1; j <= t1; ++j) now += abs(a[j] - b[i + j - 1]);
		for (int j = i + t1; j <= t2; ++j)
			now += n - b[j];
		ret = min(ret, now);
	}
	return ret;
}

int main() 
{
	scanf("%d", &n);
	scanf("%s", s + 1);
	scanf("%s", t + 1);
	if (n <= 2) 
	{
		int ans = 0;
		for (int i = 1; i <= n; ++i) ans += s[i] != t[i];
		printf("%d\n", ans);
		return 0;
	}
	int tmp = calc();
	for (int i = 1; i <= n; ++i) 
		s[i] = '0' + '1' - s[i], t[i] = '0' + '1' - t[i];
	printf("%d\n", tmp + calc());
	return 0;
}