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<cstring>
#include<cmath>
using namespace std;
const int N=5010,inf=1e9+10;
int n;
char S[N],T[N];
int s[N],t[N],s1[N],t1[N];
int ss,tt,c1,c2,ret=inf;
void solve()
{
	ss=tt=1;
	int ans=0;
	memcpy(s1,s,sizeof(s1));memcpy(t1,t,sizeof(t1));
	while (1)
	{
		while (!s1[ss]) ss++;
		while (!t1[tt]) tt++;
		if (ss>n||tt>n)break;
		s1[ss]--;t1[tt]--;ans+=abs(ss-tt);
	}
	for (int i=ss;i<=n;i++) ans+=s1[i]*(n-i+1);
	for (int i=tt;i<=n;i++) ans+=t1[i]*(n-i+1);
	ret=min(ret,ans);
}
int main()
{
	scanf("%d%s%s",&n,S+1,T+1);
	for (int i=2;i<=n;i++)
	{
		if (S[i]!=S[i-1])s[i]++;
		if (T[i]!=T[i-1])t[i]++;
	}
	if (S[1]!=T[1]) s[1]++;
	s[n+1]=t[n+1]=inf;
	for (int i=0;i<=n;i++)
	{
		s[1]+=2*i;solve();s[1]-=2*i;
	}
	for (int i=0;i<=n;i++)
	{
		t[1]+=2*i;solve();t[1]-=2*i;
	}		
	cout<<ret;
}