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 <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <iostream>
#include <algorithm>
#include <vector>
#include <set>
#include <map>
#include <stack>
#include <queue>
#define rep(i,l,r) for(int i=(l);i<=(r);++i)
#define per(i,r,l) for(int i=(r);i>=(l);--i)
using namespace std;
const int N=1e5+10;
int A[N],B[N],t1,t2,ans,n;
char s[N],t[N];
int main(){
    scanf("%*d%s%s",s+1,t+1);n=strlen(s+1);int *a=A+n,*b=B+n;
    rep(i,1,2*n) a[i]=b[i]=n;
    rep(i,1,n-1) if(s[i]!=s[i+1]) a[++t1]=i;
    rep(i,1,n-1) if(t[i]!=t[i+1]) b[++t2]=i;ans=1e9;
    rep(i,-t1-1,t2+1) if((i&1)!=(s[1]==t[1])){
        int now=0;
        rep(j,min(1,1-i),max(t1,t2-i))
            now+=abs(a[j]-b[i+j]);
        ans=min(ans,now);
    }
    printf("%d\n",ans);
    return 0;
}