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 <algorithm>
#include <vector>
using namespace std;
typedef long long ll;
int n;
int w[10001];
int f[10001];
int edge[10001];
bool used[10001];
vector<int> vs;

void dfs(int v){
	used[v]=true;
	vs.push_back(w[v]);
	if(!used[edge[v]])dfs(edge[v]);
}

int main(void){
	scanf("%d",&n);
	for(int i=0;i<n;i++){
		scanf("%d",&w[i]);
		f[i]=w[i];
	}
	sort(f,f+n);
	ll res=0;
	for(int i=0;i<n;i++){
		if(f[i]==w[i])edge[i]=i;
		else edge[i]=lower_bound(f,f+n,w[i])-f;
	}
	for(int i=0;i<n;i++){
		if(edge[i]!=i){
			vs.clear();
			dfs(i);
			int all=vs[0];
			int minx=vs[0];
			for(int j=1;j<vs.size();j++){
				minx=min(minx,vs[j]);
				all+=vs[j];
			}
			res+=min(all+(vs.size()-2)*minx,all+(vs.size()+1)*f[0]+minx);
		}
	}
	printf("%d\n",res);
	return 0;
}