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 <algorithm>
#include <cstdio>
#include <iostream>
#include <map>
#include <cmath>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <vector>
#include <stdlib.h>
#include <stdio.h>
#include <bitset>
#include <cstring>
#include <deque>
#include <iomanip>
#include <limits>
#include <fstream>
using namespace std;
#define FOR(I,A,B) for(int I = (A); I < (B); ++I)
#define CLR(mat) memset(mat, 0, sizeof(mat))
typedef long long ll;
int main()
{
  ios::sync_with_stdio(false);
  cin.tie(0);
  int N;cin>>N;
  int w[N];
  int mini = 2e9;
  map<int, int> mp;
  FOR(i,0,N){
    cin>>w[i];
    mini = min(mini, w[i]);
    mp[w[i]] = 0;
  }
  int k = 0;
  for(auto& m : mp) m.second = k++;
  bool done[N];
  CLR(done);
  int ans = 0;
  FOR(i,0,N){
    if(done[i]) continue;
    int cnt = 0, now = i, mi = 2e9, sum = 0;
    // 巡回
    while(!done[now]){
      cnt++;
      done[now] = true;
      mi = min(mi, w[now]);
      sum += w[now];
      now = mp[w[now]];
    }
    if(cnt <= 1) continue;
    ans += min(sum + mi * (cnt - 2), sum + mi + mini * (cnt + 1));
  }
  cout<<ans<<endl;
}
