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>
void qsort(int a[], int left, int right) {
  int p = (left + right) / 2;
  int pivot = a[p];
  int i, temp;
  if (left < right) {
    p = left;
    pivot = a[p];
    for (i = left + 1; i <= right; i++) {
      if (a[i] < pivot) {
        p++;
        temp = a[i];
        a[i] = a[p];
        a[p] = temp;
      }
    }
    temp = a[left];
    a[left] = a[p];
    a[p] = temp;
    qsort(a, left, p - 1);
    qsort(a, p + 1, right);
  }
}
int search(int key, int a[], int n) {
  int i;
  for (i = 0; i < n; i++) {
    if (key == a[i]) return i;
  }
  return i;
}
int isVisited(int v[], int n) {
  int i;
  for (i = 0; i < n; i++) {
    if (v[i] != 1) return -1;
  }
  return 1;
}
int min(int a, int b) {
  if (a < b) return a;
  return b;
}
int main(void) {
  int n, i, j, minW, cost = 0, w;
  int a[1000], b[1000], v[1000];
  scanf("%d", &n);
  for (i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    b[i] = a[i];
    v[i] = 0;
  }
  qsort(b, 0, n - 1);
  for (i = 0; i < n; i++) {
    int x = i;
    int sum = 0;
    int count = 0;
    minW = 1000;
    while (v[x] == 0) {
      v[x] = 1;
      sum += a[x];
      count++;
      if (minW > a[x]) minW = a[x];
      x = search(a[x], b, n);
    }
    if (count < 2) continue;
    cost += min(b[0] * (count + 1) + sum + minW, sum + (count - 2) * minW);
    if (isVisited(v, n) == 1) break;
  }
  printf("%d\n", cost);
  return 0;
}
