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 <iostream>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 5010;
int Ws[maxn], Wt[maxn], ns = 0, nt = 0;
int vs[maxn], vt[maxn], n;
int get(int x)
{
	int sum = 0;
	int j = 1;
	for(int i = 1; i <= x; i ++)
	{
		if(j > nt)
			sum += n;
		else
			sum += Wt[j];
		j ++;
	}
	for(int i = 1; i <= ns; i ++)
	{
		if(j > nt)
			sum += n - Ws[i];
		else
		{
			sum += abs(Wt[j] - Ws[i]);
			j ++;
		}
	}
	for(; j <= nt; j ++)
		sum += abs(n - Wt[j]);
	return sum;
}
int go(int x)
{
	int sum = 0;
	int j = 1;
	for(int i = 1; i <= x; i ++)
	{
		if(i > ns)
			sum += n;
		else
			sum += Ws[i];
	}
	for(int i = x + 1; i <= ns; i ++)
	{
		if(j > nt)
			sum += n - Ws[i];
		else
		{
			sum += abs(Wt[j] - Ws[i]);
			j ++;
		}
	}
	for(; j <= nt; j ++)
		sum += n - Wt[j];
	return sum;
}
int main()
{
	std::ios::sync_with_stdio(false);
	cin >> n;
	for(int i = 1; i <= n; i ++)
	{
		char a;
		cin >> a;
		vs[i] = a - '0';
	}
	for(int i = 1; i <= n; i ++)
	{
		char a;
		cin >> a;
		vt[i] = a - '0';
	}
	if(n == 1)
	{
		cout << (vs[1] ^ vt[1]) << endl;
		return 0;
	}
	for(int i = 1; i < n; i ++)
	{
		if((vs[i] ^ vs[i + 1]) == 0)
		{
			ns ++;
			Ws[ns] = i;
		}
		if((vt[i] ^ vt[i + 1]) == 0)
		{
			nt ++;
			Wt[nt] = i;
		}
	}
	if(ns > nt)
	{
		for(int i = 1; i <= ns; i ++)
			swap(Ws[i], Wt[i]);
		swap(ns, nt);
	}
	/*
	cout << ns << " " << nt << endl;
	for(int i = 1; i <= ns; i ++)
		cout << Ws[i] << " " ;
	cout << endl;
	for(int i = 1; i <= nt; i ++)
		cout << Wt[i] << " " ;
	cout << endl;
	*/
	int asd = (vs[1] ^ vt[1]);
	//cout << asd << endl;
	int ans = maxn * maxn;
	for(int i = asd; i + ns <= nt * 2 + 3; i += 2)
		ans = min(ans, get(i));
	//cout << ans << endl;
	for(int i = asd; i <= nt * 2 + 3; i += 2)
		ans = min(ans, go(i));
	cout << ans << endl;
	return 0;
}