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 <algorithm>
#include <cstring>
#include <cmath>
#include <queue>
#include <vector>
#include <map>
#include <set>
#define MAXN 5005
#define P 998244353
#define INF 0x3f3f3f3f
#define rint register int
#define LL long long
#define LD long double
using namespace std;

int n, ans=INF, t1, t2, p[MAXN], q[MAXN];
char s[MAXN], t[MAXN];

int solve(int x)
{
    int sum=0;
    for(rint i=1; i<x; ++i) sum+=p[i];
    for(rint i=1; i<=t2; ++i, ++x)
    {
        if(x<=0) sum+=q[i];
        else if(x<=t1) sum+=abs(p[x]-q[i]);
        else sum+=n-q[i];
    }
    for(rint i=max(x, 1); i<=t1; ++i) sum+=n-p[i];
    return sum;
}

int main()
{
    scanf("%d%s%s", &n, s+1, t+1);
    for(rint i=1; i<n; ++i)
        if(s[i]!=s[i+1]) p[++t1]=i;
    for(rint i=1; i<n; ++i)
        if(t[i]!=t[i+1]) q[++t2]=i;
    if(!t1 && !t2 && s[1]!=t[1]) {printf("%d\n", n); return 0;}
    int tag=(s[1]==t[1]);
    for(rint i=-t2-1; i<=t1+1; ++i)
        if((i+INF-tag)&1) ans=min(ans, solve(i));
    printf("%d\n", ans);
    return 0;
}