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>
using namespace std;

void doc(){
    freopen("OB.inp","r", stdin);
    freopen("OB.out","w", stdout);
}

typedef pair<int, int> ii;

void solve(){
    int n;
    cin >> n;
    int a[n + 1], d[(int)1e5], b[n + 1];
    for(int i = 1; i <= n; i++){
        cin >> a[i];
        d[a[i]] = i;
        b[i] = a[i];
       // cout << a[i] << " " << i << " " << d[10] << '\n';
    }

    bool KT[n + 1];
    memset(KT, true, sizeof(KT));
    sort(b + 1, b + n + 1);
    int res = 0;
   // cerr << "PK";
    for(int i = 1; i <= n; i++){
        int cnt = 0, sum = 0, m = 1e5, t = i;
        while(KT[t]){
            m = min(m, a[t]);
            sum += a[t];
            cnt++;
            KT[t] = false;
           // cout << t << " ";
            t = d[b[t]];
           // cout << t << "*" << '\n';
        }

        if (cnt >= 2) res += min(sum + m + b[1] * (cnt + 1), sum + m * (cnt - 2));
        // cout << sum << " " << m << " " << cnt << " " <<  sum + m + b[0] * (cnt + 1) << '\n';
    }
    cout << res << '\n';
}

int main(){
//    ios_base::sync_with_stdio(0);cin.tie(nullptr);cout.tie(nullptr);
 //   doc();

    solve();
}

