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;
const int Maxn=5005;
int n;
char s[Maxn],t[Maxn];
int pool1[2*Maxn],pool2[2*Maxn],*a=pool1+Maxn,*b=pool2+Maxn;
int main(){
    cin>>n>>(s+1)>>(t+1);
    int cnta=0,cntb=0,ans=INT_MAX;
    if(s[1]!=t[1])a[++cnta]=0;
    for(int i=1;i<n;i++)
        if(s[i]!=s[i+1])
            a[++cnta]=i;
    for(int i=1;i<n;i++)
        if(t[i]!=t[i+1])
            b[++cntb]=i;
    for(int i=cnta+1;i<=n;i++)a[i]=n;
    for(int i=cntb+1;i<=n;i++)b[i]=n;
    for(int i=-cnta;i<=cntb;i++){
        if(i&1)continue;
        int sum=0;
        for(int delta=-n;delta<=n;delta++){
            int pos=i+delta;
            if(pos>n)sum+=n-b[delta];
            else if(pos<-n)sum+=b[delta];
            else sum+=abs(a[pos]-b[delta]);
        }
        ans=min(ans,sum);
    }
    cout<<ans<<"\n";
    return 0;
}