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 sorting(vector<short> v) {
  unsigned char size_ = v.size();
  map<vector<short>, int> d;
  vector<short> sorted(v.begin(), v.end());
  sort(v.begin(), v.end());
  queue<pair<vector<short>, int> > que;
  que.push(make_pair(v, 1));
  d[v] = 1;
  while (!que.empty()) {
    vector<short> v1 = que.front().first;
    int cost = que.front().second;
    que.pop();
    for (char i = 0; i < size_; i++) {
      for (char j = i + 1; j < size_; j++) {
        vector<short> v2(v1.begin(), v1.end());
        int cost2 = cost + v1[i] + v1[j];
        swap(v2[i], v2[j]);
        if (d[v2] == 0 || d[v2] > cost2) {
          d[v2] = cost2;
          que.push(make_pair(v2, cost2));
        }
      }
    }
  }
  return d[sorted] - 1;
}
int main() {
  vector<short> v;
  int n;
  scanf("%d", &n);
  v = vector<short>(n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &v[i]);
  }
  printf("%d\n", sorting(v));
  return 0;
}
