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;
	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];
		chk[i] = NOT_CHECK;
	}

	qsort((void*)sortedA, n, sizeof(element), element_sort);
	// 荷物の重さから元々の位置を取得できるように値を設定
	for(i=0;i<n;i++) W[sortedA[i].val] = i;

	for(i=0;i<n;i++) {
		int p, cost;
		p = sortedA[i].idx;

		if(sortedA[i].idx == i) {
			// ソート後と同じ位置のため入れ替える必要がない
			chk[i] = CHECK;
			continue;
		}

		// 軽い荷物を使って、正しい場所に入れ替えていく
		cost = 0;
		while(!chk[p]) {
			chk[p] = CHECK;
			cost = cost + A[p];
			p = W[A[p]];
		}
		costs += cost;
	}

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