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 <algorithm>
#include <cstdio>
#include <vector>
template <typename T> constexpr T my_abs(const T &x) { return x < 0 ? -x : x; }
char s[5005], t[5005];
int main()
{
	// freopen("AGC030-E.in", "r", stdin);
	int n;
	scanf("%d%s%s", &n, s, t);
	auto init = [&] (char *str)
	{
		std::vector<int> res;
		for (int i = 1; i < n; i++)
		{
			if (str[i] != str[i - 1])
				res.push_back(i);
		}
		return res;
	};
	auto vec_s = init(s), vec_t = init(t);
	auto get = [&] (std::vector<int> &vec, int idx)
	{
		if (idx < 0)
			return 0;
		if (idx >= vec.size())
			return n;
		return vec[idx];
	};
	int ans = 1e9, a = vec_s.size(), b = vec_t.size();
	for (int off = -a - 1; off <= b + 1; off++)
	{
		if ((off & 1) == (s[0] == t[0]))
			continue;
		int res = 0;
		for (int i = std::min(-off, 0); i < std::max(a, b - off); i++)
			res += my_abs(get(vec_s, i) - get(vec_t, i + off));
		ans = std::min(ans, res);
	}
	printf("%d\n", ans);
	return 0;
}
