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 <algorithm>
#include <vector>

using namespace std;

int main() {
	int n, cost1 = 0, cost2 = 0;
	cin >> n;
	
	int arr1[n], arr2[n], sorted[n];
	bool visited[n] = { false };
	for (int i = 0; i < n; i++) {
		cin >> arr1[i];
		arr2[i] = sorted[i] = arr1[i];
	}
	sort(sorted, sorted + n);

	vector<pair<int, int> > vec;
	int mini = 0, s1, s2;
	while (mini < n) {
		for (int i = 0; i < n; i++) {
			if (arr1[i] == sorted[mini]) {
				s1 = i;
				break;
			}
		}
		for (int i = 0; i < n; i++) {
			if (arr1[i] == sorted[s1]) {
				s2 = i;
				break;
			}
		}

		if (s1 != s2) {
			visited[mini] = visited[s1] = true;
			cost1 += arr1[s1] + arr1[s2];
			vec.push_back(make_pair(s1, s2));
			swap(arr1[s1], arr1[s2]);
		}
		else {
			mini++;
		}
	}

	s1 = s2 = -1;
	for (int i = 0; i < n; i++) {
		if (s1 == -1 && visited[i] == true) {
			s1 = i;
		}
		if (s2 == -1 && visited[i] == false) {
			s2 = i;
		}
	}

	if (s1 != -1 && s2 != -1) {
		for (int i = 0; i < n; i++) {
			if (arr2[i] == sorted[s1]) {
				s1 = i;
			}
			if (arr2[i] == sorted[s2]) {
				s2 = i;
			}
		}

		cost2 += (arr2[s1] + arr2[s2]) * 2;
		swap(arr2[s1], arr2[s2]);

		int size = vec.size();
		for (int i = 0; i < size; i++) {
			cost2 += arr2[vec[i].first] + arr2[vec[i].second];
			swap(arr2[vec[i].first], arr2[vec[i].second]);
		}

		cout << min(cost1, cost2) << endl;
	}
	else {
		cout << cost1 << endl;
	}

	return 0;
}