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;
int main() {
  int n, w, l;
  vector<pair<int, pair<int, int> > > a;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> w;
    a.push_back(make_pair(w, make_pair(i, 0)));
  }
  sort(a.begin(), a.end());
  for (int i = 0; i < n; i++) {
    (a[i].second).second = i;
    l = a[i].first;
    a[i].first = (a[i].second).first;
    (a[i].second).first = l;
  }
  sort(a.begin(), a.end());
  int mi = a[0].first;
  bool b[1010] = {0};
  int ans = 0;
  for (int i = 0; i < n; i++) {
    if (b[i] != 0) continue;
    b[i] = 1;
    int j = i, k = 1, sum = (a[i].second).first, mini = (a[i].second).first;
    while (1) {
      if (i == (a[j].second).second) break;
      j = (a[j].second).second;
      b[j] = 1;
      sum = sum + (a[j].second).first;
      k++;
      if (mini > (a[j].second).first) mini = (a[j].second).first;
    }
    if (k > 1) ans = ans + min(sum + (k - 2) * mini, sum + (k + 1) * mi + mini);
  }
  cout << ans << endl;
  return 0;
}
