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 = INT_MAX;
const int MOD = (int)1e9 + 7;
const double EPS = 1e-12;
const int dx[] = {-1, -1, -1, 0, 1, 1, 1, 0},
          dy[] = {-1, 0, 1, 1, 1, 0, -1, -1};
using namespace std;
signed main() {
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  int flag[1000] = {};
  int where[10010] = {};
  vector<int> a(n), sorted_a;
  for (int i = 0; i < (n); i++) cin >> a[i], where[a[i]] = i;
  sorted_a = a;
  sort(sorted_a.begin(), sorted_a.end());
  vector<vector<int>> cost_list;
  for (int i = 0; i < (n); i++) {
    if (flag[i]) continue;
    int now = i;
    vector<int> cost;
    for (; !flag[now];) {
      cost.push_back(a[now]);
      flag[now] = 1;
      now = where[sorted_a[now]];
    }
    cost_list.push_back(cost);
  }
  int ans = 0, min_a = sorted_a[0];
  for (auto e : cost_list) {
    int r1 = accumulate(e.begin(), e.end(), 0) +
             (e.size() - 2) * *min_element(e.begin(), e.end());
    int r2 = accumulate(e.begin(), e.end(), 0) +
             *min_element(e.begin(), e.end()) + (e.size() - 1) * min_a;
    ans += min(r1, r2);
  }
  cout << ans << endl;
  return 0;
}
