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<functional>
#include<algorithm>
#include<iterator>
#include<cstring>
#include<cstdio>
using namespace std;
int n;
int c[11111];
char s[11111],t[11111];
int werken(int cs,int ct)
{
	int tmp=0;
	for(int i=0;i<n;++i)
	{
		++c[cs-ct+5555];
		tmp+=cs-ct+5555;
		if(s[i]!=s[i+1])++cs;
		if(t[i]!=t[i+1])++ct;
	}
	int cu=n,cd=0,res=2147483647;
	for(int i=0;i<11111;++i)
	{
		if((i&1)&&tmp<res)res=tmp;
		cu-=c[i];
		cd+=c[i];
		tmp-=cu;
		tmp+=cd;
	}
	return res;
}
int main()
{
	scanf("%d",&n);
	scanf("%s%s",s,t);
	if(s[0]==t[0])printf("%d",werken(0,0));
	else printf("%d",werken(0,1));
	return 0;
}