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 rep(i,a,b) for (int i = (a); i <= (b); ++ i)
#define rrp(i,a,b) for (int i = (a); i >= (b); -- i)
#define gc() getchar()
#define fir first
#define sec second
typedef pair<int,int> pii;
typedef long double db;
typedef long long ll;
template <typename tp>
inline void read(tp& x) {
  x = 0; char tmp; bool key = 0;
  for (tmp = gc(); !isdigit(tmp); tmp = gc())
    key = (tmp == '-');
  for (; isdigit(tmp); tmp = gc())
    x = (x << 3) + (x << 1) + (tmp ^ '0');
  if (key) x = -x;
}
template <typename tp>
inline void ckmn(tp& x,tp y) {
  x = x < y ? x : y;
}
template <typename tp>
inline void ckmx(tp& x,tp y) {
  x = x < y ? y : x;
}

const int N = 5010, INF = 0x3f3f3f3f;
char s[N], t[N];
int n,ans,pla[N << 1],cnt,rec[N << 1];
int solve(int p) {
  int tmp = 0;
  rep (i, 1, p) pla[++tmp] = 0;
  int ret = 0;
  if (p >= 0) pla[++tmp] = 0;
  else ++ p;
  for (int i = 1; i < n; ++ i) {
    if (s[i] != s[i+1]) {
      if (p < 0) ++ p, ret += i;
      else pla[++tmp] = i;
    }
  }
  while (tmp < cnt) pla[++tmp] = n;
  rep (i, 1, tmp) {
    if (i <= cnt) ret += abs(pla[i] - rec[i]);
    else ret += abs(pla[i] - n);
  }
  return ret;
}
int main() {
  read(n);
  scanf("%s", s+1);
  scanf("%s", t+1);
  rec[++cnt] = 0;
  rep (i, 1, n-1) if (t[i] != t[i+1])
    rec[++ cnt] = i;
  while (cnt < n) rec[++cnt] = n;
  int key = 0;
  if (s[1] != t[1]) key = 1;
  ans = INF;
  rep (i, -n-1, n+1) if ((abs(i)&1) == key)
    ans = min(ans, solve(i));
  printf("%d\n", ans);
  return 0;
}
