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;
#define ri register int
const int N=5e3+5;
char s[N],t[N];
int a[N],b[N];
int main(){
	ri n,ns,nt,i,j,res,ans;
	scanf("%d%s%s",&n,s+1,t+1);
	ns=nt=0;
	for(i=1;i<n;++i)
		if(s[i]!=s[i+1])
			a[++ns]=i;
	for(i=1;i<n;++i)
		if(t[i]!=t[i+1])
			b[++nt]=i;
	ans=2e9;
	for(i=-max(ns,nt)-1;i<=max(ns,nt)+1;++i)
		if((i&1)^(s[1]==t[1])){
			res=0;
			for(j=min(1,1-i);j<=max(ns,nt-i);++j)
				res+=abs((j<0?0:(j>ns?n:a[j]))-(j+i<0?0:(j+i>nt?n:b[j+i])));
			ans=min(ans,res);
		}
	printf("%d\n",ans);
	return 0;
}