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 <cstring>
#include <cstdlib>
#include <sstream>
#include <stack>
#include <queue>
#include <list>
#include <stdio.h>
#include <algorithm>
#define MAX 100000
using namespace std;
typedef long long ll;
ll ans;
void solve(int a[],int n)
{
	int b[1000],t[10005],vmin,vmax=10000;
	bool vis[1000];
	for(int i=0;i<n;i++)
	{
		b[i]=a[i];
		vis[i]=false;
	}
	sort(b,b+n);
	vmin=b[0];
	for(int i=0;i<n;i++)
	{
		t[b[i]]=i;//得出移动够各个元素（值不同）应该在数组中位置 
	}
	for(int i=0;i<n;i++)
	{
		int cur=i,m=vmax,sum=0,num=0;//查找新的环 
		while(1)//开始搜索下一个环 
		{ 
			vis[cur]=true;//将这个元素加入当前环 
			num++;//这个环的个数加1 
			sum+=a[cur];//环的weight加1 
			m=min(a[cur],m);//更新最小值，最终得到该环的最小值 
			cur=t[a[cur]];//得到将要移动到的位置 
			if(vis[cur]) break;//将要移动到的位置的元素已经在环中，环已经闭合，一个环已经找到 
		}
		ans+=min(sum+(num-2)*m,sum+m+(num+1)*vmin);//如果已经在某个环中或者根本不需要移动，那么成本为0，不影响结果 
	}
}
int main()
{	
	int n,a[1000];
	cin>>n;
	for(int i=0;i<n;i++)
	{
		scanf("%d",&a[i]);
	}
	solve(a,n);
	cout<<ans<<endl;
	return 0;
}
