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;
int n;
void get(int *p, int &pc, char *s) {
    if (s[1] == '0') p[++pc] = 0;
    p[++pc] = 0;
    for (int i = 1; i < n; ++i)
        if (s[i] != s[i + 1]) p[++pc] = i;
    if (s[n] == '0') p[++pc] = n;
    p[++pc] = n;
}

char s[5003], t[5003];
int p[5003], q[50003], pc, qc;
int res = 1000000000;
int qry(int *p, int pc, int pos) {
    return pos < 1 ? 0 : (pos > pc ? n : p[pos]);
}
int solve(int x) {
    int rt = 0;
    for (int i = min(1 - x, 1), en = max(qc - x, pc); i <= en; ++i)
        rt += abs(qry(p, pc, i) - qry(q, qc, i + x));
    return rt;
}
int main() {
    scanf("%d%s%s", &n, s + 1, t + 1);
    get(p, pc, s), get(q, qc, t);
    for (int i = -pc; i <= qc; i += 2) res = min(res, solve(i));
    printf("%d\n", res);
    return 0;
}
