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 MN 5000
inline int read()
{
    int x=0,f=1;char c=getchar();
    while(c<'0'||c>'9'){if(c=='-') f=-1;c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
    return x*f;
}
int n,ans=1e9;
vector<int> v1,v2;
char s[MN+5],t[MN+5];
inline int Abs(int x){return x>0?x:-x;}
int main()
{
    n=read();scanf("%s%s",s+1,t+1);
    for(int i=2;i<=n;++i) if(s[i]!=s[i-1]) v1.push_back(i-1);
    for(int i=2;i<=n;++i) if(t[i]!=t[i-1]) v2.push_back(i-1);
    if(n<=2)
    {
        int res=0;
        for(int i=1;i<=n;++i) res+=(s[i]!=t[i]);
        return 0*printf("%d",res);
    }
    int s1=v1.size(),s2=v2.size();
    //for(int i=0;i<s1;++i) printf("%d ",v1[i]);cout<<endl;
    //for(int i=0;i<s2;++i) printf("%d ",v2[i]);cout<<endl;
    for(int i=0-(s[v1[0]]!=t[v2[0]])-s1-(s1&1);i<=s2+1;i+=2)
    {
        int res=0;
        for(register int j=0;j<s1;++j)
        {
            int t=j+i,pos=t<0?0:(t>=s2?n:v2[t]);
            res+=Abs(v1[j]-pos);
        }
        //cout<<"preres="<<res<<endl;
        for(register int j=0;j<s2&&j<i;++j) res+=v2[j];
        for(register int j=s2-1;~j&&j-i>=s1;--j) res+=n-v2[j];
        ans=min(ans,res);
        //cout<<i<<" "<<ans<<endl;
    }
    cout<<ans;
    return 0;
}