You are given $n$ integers $w_i (i = 0, 1, ..., n-1)$ to be sorted in ascending order. You can swap two integers $w_i$ and $w_j$. Each swap operation has a cost, which is the sum of the two integers $w_i + w_j$. You can perform the operations any number of times.

Write a program which reports the minimal total cost to sort the given integers.

Constraints

* $1 \leq n \leq 1,000$
* $0 \leq w_i\leq 10^4$
* $w_i$ are all different

Input

In the first line, an integer $n$ is given. In the second line, $n$ integers $w_i (i = 0, 1, 2, ... n-1)$ separated by space characters are given.

Output

Print the minimal cost in a line.

Examples

Input

5
1 5 3 4 2


Output

7


Input

4
4 3 2 1


Output

10
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e3 + 5;
const int inf = 1e9 + 5;
int a[maxn], b[maxn], s, n, p[maxn];
bool v[maxn];
long long solve() {
  long long an = 0, ans = 0;
  int mi = inf;
  for (int i = 0; i < n; i++) p[a[i]] = i;
  for (int i = 0; i < n; i++) {
    if (a[i] == i)
      v[i] = 1;
    else
      v[i] = 0;
  }
  for (int i = 0; i < n; i++) {
    int k = i;
    an = 0;
    mi = inf;
    if (v[k]) continue;
    int r = 1;
    while (1) {
      if (v[k]) break;
      v[k] = 1, ++r;
      an += a[k] + 1;
      mi = min(mi, a[k] + 1);
      k = a[p[k]];
    }
    if (r >= 3)
      ans += min(an + (n - 2) * mi,
                 an + (n - 2) * mi + 2 * (mi + s) - (n - 1) * (mi - s));
    else if (r == 2)
      ans += an;
  }
  return ans;
}
int main() {
  scanf("%d", &n);
  s = inf;
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]), --a[i];
    s = min(s, a[i]);
  }
  long long ans = solve();
  cout << ans << endl;
  return 0;
}
