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 <string>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
int main() {
	int n; string s, t;
	cin >> n >> s >> t;
	vector<int> sps = { 0 }, spt = { 0 };
	for (int i = 1; i < n; ++i) {
		if (s[i - 1] != s[i]) {
			sps.push_back(i);
		}
		if (t[i - 1] != t[i]) {
			spt.push_back(i);
		}
	}
	sps.push_back(n);
	spt.push_back(n);
	int parity = (s[0] == t[0] ? 0 : 1);
	int ls = sps.size(), lt = spt.size(), ans = (1 << 30);
	for (int i = -ls + 1; i < lt; ++i) {
		if (abs(i) % 2 != parity) continue;
		int sum = 0;
		for (int j = -lt + 1; j < ls + lt; ++j) {
			int ps = j - i;
			if (ps < 0) ps = 0;
			else if (ps >= ls) ps = n;
			else ps = sps[ps];
			int pt = j;
			if (pt < 0) pt = 0;
			else if (pt >= lt) pt = n;
			else pt = spt[pt];
			sum += abs(ps - pt);
		}
		ans = min(ans, sum);
	}
	cout << ans << endl;
	return 0;
}