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>
const int INF = 1e9;
using namespace std;
bool stable = true;
int main() {
  int n;
  int cost = 0;
  cin >> n;
  vector<int> a(n);
  vector<bool> miss(n);
  for (int i = 0; i < (int)(n); i++) cin >> a[i];
  auto b = a;
  sort((a).begin(), (a).end());
  for (int i = 0; i < (int)(n); i++) {
    miss[i] = (a[i] != b[i]);
  }
  for (int i = 0; i < (int)(n); i++) {
    for (int i = 0; i < (int)(n); i++) {
      if (miss[n - 1 - i]) {
        int replace = find((b).begin(), (b).end(), a[n - 1 - i]) - b.begin();
        if (b[n - 1 - i] == a[replace]) {
          swap(b[n - 1 - i], b[replace]);
          cost += b[n - 1 - i] + b[replace];
          miss[n - 1 - i] = false;
          miss[replace] = false;
        }
      }
    }
    {
      if (miss[n - 1 - i]) {
        int replace = find((b).begin(), (b).end(), a[n - 1 - i]) - b.begin();
        miss[n - 1 - i] = false;
        if (b[n - 1 - i] == a[replace]) miss[replace] = false;
        swap(b[n - 1 - i], b[replace]);
        cost += b[n - 1 - i] + b[replace];
      }
    }
  }
  cout << cost << endl;
  return 0;
}
