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>
int array[1001];
int sorted[1001];
int swap_limit = 0;
int answer = 0;
void swap(int* a, int* b) {
  int tmp = *a;
  *a = *b;
  *b = tmp;
}
void dfs(int n, int deep, int score) {
  int i, j, flag = 1;
  for (i = 0; i < n; i++) {
    if (array[i] == sorted[i]) {
      continue;
    }
    for (j = i + 1; j < n; j++) {
      if (array[j] == sorted[j]) {
        continue;
      }
      {
        int next_score = score + array[i] + array[j];
        flag = 0;
        if (next_score >= answer) {
          continue;
        } else if (deep + 1 > swap_limit) {
          continue;
        } else {
          swap(&array[i], &array[j]);
          dfs(n, deep + 1, next_score);
          swap(&array[i], &array[j]);
        }
      }
    }
  }
  if (flag) {
    if (answer > score) {
      answer = score;
    }
    if (swap_limit > deep) {
      swap_limit > deep;
    }
  }
}
int main(int argc, char* argv[]) {
  int i, j, n;
  scanf("%d", &n);
  for (i = 0; i < n; i++) {
    scanf("%d", &array[i]);
    sorted[i] = array[i];
  }
  for (i = 0; i < n; i++) {
    int min = sorted[i];
    int min_pos = i;
    for (j = i + 1; j < n; j++) {
      if (min > sorted[j]) {
        min = sorted[j];
        min_pos = j;
      }
    }
    if (min_pos != i) {
      swap(&sorted[i], &sorted[min_pos]);
      swap_limit++;
      answer += sorted[i] + sorted[min_pos];
    }
  }
  dfs(n, 0, 0);
  printf("%d\n", answer);
  return 0;
}
