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>
using namespace std;

typedef pair<int, int> pii;
typedef long long ll;
typedef vector<int> vi;

#define pb push_back
#define eb emplace_back
#define mp make_pair
#define fi first
#define se second
#define rep(i,n) rep2(i,0,n)
#define rep2(i,m,n) for(int i=m;i<(n);i++)
#define ALL(c) (c).begin(),(c).end()
#define dump(x) cout << #x << " = " << (x) << endl
constexpr ll TEN(int n) { return (n == 0) ? 1 : 10 * TEN(n-1); }

template<class T, class U>
ostream& operator<<(ostream& os, const pair<T, U>& p) {
	os<<"("<<p.first<<","<<p.second<<")";
	return os;
}

template<class T>
ostream& operator<<(ostream& os, const vector<T>& v) {
	os<<"{";
	rep(i, v.size()) {
		if (i) os<<",";
		os<<v[i];
	}
	os<<"}";
	return os;
}

int main() {
	int N; string s, t;
	cin >> N >> s >> t;

	if (s == t) {
		puts("0");
		return 0;
	}

	vector<pii> v[2];

	rep(i, N-1) {
		if (s[i] != s[i+1]) {
			v[0].eb(s[i]-'0', i);
		}
		if (t[i] != t[i+1]) {
			v[1].eb(t[i]-'0', i);
		}
	}

	if (v[0].size() == 0) {
		cout << N << endl;
		return 0;
	}

	ll ans = TEN(18);

	for (int d = -N; d <= N; ++d) {
		deque<pii> s[2];
		rep(t, 2) {
			for (auto p : v[t]) {
				s[t].eb(p);
			}		
		}

		if (d < 0) {
			rep(i, -d) {
				s[1].push_front(mp(-1, -1));
			}
		} else {
			rep(i, d) {
				s[0].push_front(mp(-1, -1));
			}
		}
		while (s[0].size() < s[1].size()) {
			s[0].eb(mp(-1, N-1));
		}
		while (s[0].size() > s[1].size()) {
			s[1].eb(mp(-1, N-1));
		}

		bool ok = true;
		ll dis = 0;

		rep(i, s[0].size()) {
			if (s[0][i].fi != -1 && s[1][i].fi != -1 && s[0][i].fi != s[1][i].fi) {
				ok = false;
			}
			dis += abs(s[0][i].se - s[1][i].se);
		}

		if (ok) {
			ans = min(ans, dis);
		}
	}

	cout << ans << endl;

	return 0;
}