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<cmath>
#include<vector>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
const int inf=1e9;
char s[5010],t[5010];
int s1,s2,n,ans;
vector<int> a,b;
void solve(vector<int> &a,char *s) {for(int i=1;i<n;i++) if(s[i]!=s[i+1]) a.push_back(i);}
int main()
{
	scanf("%d%s%s",&n,s+1,t+1);
	solve(a,s);s1=a.size();
	solve(b,t);s2=b.size();
	ans=inf;
	for(int k=-s1-1;k<=s2+1;k++)
		if((k+(s[1]==t[1]))&1)
		{
			int tot=0;
			for(int i=min(-k,0);i<=max(s1,s2-k);i++)
			{
				int j=i+k,px,py;
				px=i<0?0:(i>=s1?n:a[i]);
				py=j<0?0:(j>=s2?n:b[j]);
				tot+=abs(px-py);
			}
			ans=min(ans,tot);
		}
	printf("%d",ans);
}