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>
#define rep(i,x,y) for (int i=(x);i<=(y);i++)
#define VI vector<int>
using namespace std;
const int N=10005,inf=1e9;
int n,sz_a,sz_b,ans; VI a,b; char s[N],t[N];
void work(VI &a,char *s){
	rep (i,1,n-1) if (s[i]!=s[i+1]) a.push_back(i);
}
int main(){
	scanf("%d%s%s",&n,s+1,t+1);
	work(a,s); sz_a=a.size();
	work(b,t); sz_b=b.size();
	ans=inf;
	rep (dif,-sz_a-1,sz_b+1)
		if ((dif+(s[1]==t[1]))&1){
			int now=0;
			rep (i,min(-dif,0),max(sz_a,sz_b-dif)-1){
				int j=i+dif,px,py;
				px=i<0?0:i>=sz_a?n:a[i],py=j<0?0:j>=sz_b?n:b[j];
				now+=abs(px-py);
			}
			ans=min(ans,now);
		}
	printf("%d",ans);
	return 0;
}