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 <cstdio>
#include <utility>
#include <unordered_map>
#include <vector>
#include <algorithm>

int main(){
    int n;
    std::scanf("%d", &n);
    std::vector<int> a(n, 0);
    std::vector<int> b(n, 0);
    std::unordered_map<int, int> hash;
    for(int i = 0; i < n; ++i){
        std::scanf("%d", &a[i]);
        b[i] = a[i];
        hash.insert(std::make_pair(a[i], i));
    }
    std::sort(b.begin(), b.end());

    int totalCost = 0;
    int i = 0;
    int s = b[0];
    for(int i = 0; i < n; ++i){
        if(a[i] == b[i]) continue;
        int sum = 0;
        int op = 0;
        while(1){
            ++op;
            std::pair<int, int> minp;
            minp.first = hash.find(b[i])->second;
            minp.second = hash.find(b[minp.first])->second;
            sum += a[minp.second];
            std::swap(a[minp.first], a[minp.second]);
            hash.find(b[i])->second = minp.second;
            hash.find(b[minp.first])->second = minp.first;
            if(a[i] == b[i]) break;
        }
        totalCost += std::min(sum + op * b[i], sum + (op + 2) * s + 2 * b[i]);
    }
    std::printf("%d\n", totalCost);
    return 0;
}
