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, min_w = 10000;
  cin >> n;
  vector<int> W(n), W_origin(n);
  deque<int> w_cycle;
  for (size_t i = 0; i < n; i++) {
    cin >> W[i];
    min_w = min(min_w, W[i]);
  }

  copy(W.begin(), W.end(), W_origin.begin());
  sort(W.begin(), W.end());

  int cost = 0;
  for (size_t i = 0; i < n; i++) {
    if (W[i] == -1) continue;
    int idx = i;
    while (true) {
      auto correspond_w = find(W.begin(), W.end(), W_origin[idx]);
      if (correspond_w == W.end()) break;
      w_cycle.push_back(*correspond_w);

      idx = distance(W.begin(), correspond_w);
      *correspond_w = -1;
    }

    int cost_inter, cost_ext;
    auto min_w_cycle = min_element(w_cycle.begin(), w_cycle.end());

    cost_inter = accumulate(w_cycle.begin(), w_cycle.end(), 0) +
                 *min_w_cycle * (w_cycle.size() - 2);
    cost_ext = accumulate(w_cycle.begin(), w_cycle.end(), 0) + *min_w_cycle +
               min_w * (w_cycle.size() + 1);

    cost += min(cost_ext, cost_inter);

    w_cycle.clear();
  }

  cout << cost << endl;
}
