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 <stdio.h>
#include <utility>
#include <algorithm>
#include <bitset>
using namespace std;
typedef pair<int,int> pi;
#define F first
#define S second
#define MP make_pair
const int N=1e3+10;
bitset<N> went;
int bot,tot,a[N],t;
pi b[N];
void dfs(int n){
    if(went[n])return ;
    went[n]=true;
    t++;
    tot+=b[a[n]].F;
    bot=min(bot,b[a[n]].F);
    dfs(a[n]);
    return ;
}
int main(){
    int n,low=1e5+10;;
    long long int ans=0;
    scanf("%d",&n);
    for(int i=0;i<n;i++){
        scanf("%d",&b[i].F);
        low=min(low,b[i].F);
        b[i].S=i;
    }
    sort(b,b+n);
    for(int i=0;i<n;i++)a[b[i].S]=i;
    went.reset();
    for(int i=0;i<n;i++)if(!went[i]&&a[i]!=i){
        tot=0;
        bot=1e5+10;
        t=0;
        dfs(i);
        ans+=min(tot+(t+1)*low+bot,tot+(t-2)*bot);
    }
    printf("%lld\n",ans);
}
