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 <stdio.h>
#include <stdlib.h>

#define NOT_CHECK 0
#define CHECK 1

typedef struct element_ {
	int idx, val;
} element;

int element_sort(void *a, void *b) {
	return (*(element*)a).val - (*(element*)b).val;
}

int main(void) {
	int i, n, costs, m;
	int A[1005];
	int W[10005];
	int chk[1005];
	element sortedA[1005];

	scanf("%d\n", &n);
	for(i=0;i<n;i++) {
		scanf("%d\n", &A[i]);
		sortedA[i].idx = i;
		sortedA[i].val = A[i];
		W[A[i]] = i;
		chk[i] = NOT_CHECK;
	}

	qsort((void*)sortedA, n, sizeof(element), element_sort);

	m = sortedA[0].val;
	for(i=0;i<n;i++) {
		int p, cost, lt, cnt, t;
		p = i;

		t = 0;
		lt = 10005;
		while(chk[p]==NOT_CHECK) {
			chk[p] = CHECK;
			t += A[p];
			cnt++;
			if(lt > A[p]) lt = A[p];
			p = W[sortedA[p].val];
		}
		if(cnt<2) continue;	// 入れ替えが発生し始めるのはcntが2以上になってから
		
		// 入れ替え中に見つかった最小値を入れ替えに使った方がコストが低いか
		// 回り道して最小の値を使って入れ替えに使った方がコストが低いか
		cost = t+m*(cnt+1)+lt;	// lt -> m ・・ m -> A[k] ・・ m -> lt
		if(cost>t+lt*(cnt-2)) cost = t+lt*(cnt-2);
		costs += t;
	}

	printf("%d\n", costs);
	return 0;
}