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 <map>

using namespace std;
 

int solve(int *w,int n) {
	int g_m = w[0]; //整个数组中的最小值 
	 
	map<int,int> address; //元素在排序后的数组中的下标映射 
	int *tmp = new int[n];
	bool *T = new bool[n]; //标记当前位置是否被访问 
	
	int ans = 0; //最终结果 
	
	for (int i=0;i<n;i++) {
		//找出最小值 
		if (w[i] < g_m) {
			g_m = w[i];
		}
		//初始化 
		T[i] = false;
		tmp[i] = w[i];
	}
	//对临时数组排序 
	sort(tmp,tmp+n);
	//建立映射 
	for (int i=0;i<n;i++) {
		pair<int, int> pair_map(tmp[i],i);
		address.insert(pair_map);
	}
	for (int i = 0;i<n;i++) {
		if (T[i]) { //当前位置已被访问过了 
			continue;
		}
		int m = w[i]; //当前圈里的最小值 
		int cur = i;//当前所处的位置
		int S = 0; //当前所需的花费
		int count = 0; //当前圈里的元素个数
		 
		while (1) {
			count++;
			int v = w[cur];
			m = min(v,m);
			S += v;
			T[cur] = true;
			cur = address.at(v); 
			if (T[cur]) {
				break;
			}
		}
		ans += min(S + (count - 2) * m,S + m + (1+count) * g_m);
	}
	return ans;
}

int main() {
	int n;
	cin >> n;
	int *w = new int[n];
	for (int i = 0;i<n;i++) {
		cin >> w[i];
	}
	int ans = solve(w,n);
	cout << ans << endl;
	return 0;
}
