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 <iostream>
#include <cstdio>
#include <climits>
#include <cmath>
#include <algorithm>
#include <utility>
#include <vector>
#include <string>
#define MAX 1000
#define INFTY 1<<30 // 2^30
using namespace std;
typedef long long llong;

static const int VMAX = 10000;
int n, A[MAX+10], s;
int B[MAX+10], T[VMAX+10];

int solve()
{
  int ans = 0;
  bool V[MAX+10];
  for(int i = 0; i < n; i++)
  {
    B[i] = A[i];
    V[i] = false;
  }
  sort(B,B+n);
  for(int i = 0; i < n; i++) T[B[i]] = i; // ????????????????????????????????????????????????
  for(int i = 0; i < n; i++)
  {
    if(V[i]) continue;
    int cur = i; // current point
    int S = 0; // sum of mass
    int m = VMAX; // min of the cycle
    int an = 0; // the number of material the cycle have
    while(1)
    {
      V[cur] = true; //already visited
      an++;
      int v = A[cur];
      m = min(m, v);
      S += v;
      cur = T[v];
      if(V[cur]) break;
    }
    ans += min(S+(an-2)*m, S+(an+1)*s+m);
  }
  return ans;
}

int main()
{
  cin >> n;
  s = VMAX;
  for(int i = 0; i < n; i++)
  {
    cin >> A[i];
    s = min(s,A[i]);
  }
  cout << solve() << endl;
  return 0;
}