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 A[1000] = {0};
int B[1000] = {0};
int T[10000 + 1];
int n, s;
int partition(int A[], int p, int r) {
  int x, i, j, tmp;
  x = A[r];
  i = p - 1;
  for (j = p; j < r; ++j) {
    if (A[j] <= x) {
      ++i;
      tmp = A[i];
      A[i] = A[j];
      A[j] = tmp;
    }
  }
  tmp = A[i + 1];
  A[i + 1] = A[r];
  A[r] = tmp;
  return i + 1;
}
void quicksort(int A[], int p, int r) {
  int q;
  if (p < r) {
    q = partition(A, p, r);
    quicksort(A, p, q - 1);
    quicksort(A, q + 1, r);
  }
}
int solve() {
  int ans = 0;
  int i, cur, S, m, an, v;
  short V[1000] = {0};
  for (i = 0; i < n; ++i) {
    B[i] = A[i];
    V[i] = 0;
  }
  quicksort(B, 0, n - 1);
  printf("\n");
  for (i = 0; i < n; ++i) T[B[i]] = i;
  for (i = 0; i < n; ++i) {
    if (V[i]) continue;
    cur = i;
    S = 0;
    m = INT_MAX;
    an = 0;
    while (1) {
      V[cur] = 1;
      an++;
      v = A[cur];
      m = ((m) < (v) ? (m) : (v));
      S += v;
      cur = T[v];
      if (V[cur]) break;
    }
    ans +=
        ((S + (an - 2) * m) < (m + S + (an + 1) * s) ? (S + (an - 2) * m)
                                                     : (m + S + (an + 1) * s));
  }
  return ans;
}
int main(void) {
  int i;
  scanf("%d", &n);
  s = INT_MAX;
  for (i = 0; i < n; ++i) {
    scanf("%d", &A[i]);
    if (s > A[i]) s = A[i];
  }
  printf("%d\n", solve());
  return 0;
}
