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 inf 0x3f3f3f3f

using namespace std;

char s1[5005],s2[5005];
int a[5005],b[5005];

int main() {
  int n;
  scanf("%d%s%s",&n,s1+1,s2+1);
  int sz1=0,sz2=0;
  for(int i=1;i<n;i++)
    if (s1[i]!=s1[i+1]) a[++sz1]=i;
  for(int i=1;i<n;i++)
    if (s2[i]!=s2[i+1]) b[++sz2]=i;
  int ans=inf;
  for(int i=-sz1-1;i<=sz2+1;i++)
    if ((i&1)^(s1[1]==s2[1])) {
    	int s=0;
    	for(int j=min(1,1-i);j<=max(sz1,sz2-i);j++)
    	  s+=abs(((j<0)?0:((j>sz1)?n:a[j]))-((j+i<0)?0:((j+i>sz2)?n:b[j+i])));
    	ans=min(ans,s);
	}
  printf("%d\n",ans);
  return 0;
}