Algorithm

B. Shooting
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

Recently Vasya decided to improve his pistol shooting skills. Today his coach offered him the following exercise. He placed n cans in a row on a table. Cans are numbered from left to right from 11 to n. Vasya has to knock down each can exactly once to finish the exercise. He is allowed to choose the order in which he will knock the cans down.

Vasya knows that the durability of the i-th can is ai��. It means that if Vasya has already knocked x cans down and is now about to start shooting the i-th one, he will need (aix+1)(��⋅�+1) shots to knock it down. You can assume that if Vasya starts shooting the i-th can, he will be shooting it until he knocks it down.

Your task is to choose such an order of shooting so that the number of shots required to knock each of the n given cans down exactly once is minimum possible.

Input

The first line of the input contains one integer n (2n1000)(2≤�≤1000) — the number of cans.

The second line of the input contains the sequence a1,a2,,an�1,�2,…,�� (1ai1000)(1≤��≤1000), where ai�� is the durability of the i-th can.

Output

In the first line print the minimum number of shots required to knock each of the n given cans down exactly once.

In the second line print the sequence consisting of n distinct integers from 11 to n — the order of indices of cans that minimizes the number of shots required. If there are several answers, you can print any of them.

Examples
input
Copy
3
20 10 20
output
Copy
43
1 3 2
input
Copy
4
10 10 10 10
output
Copy
64
2 1 4 3
input
Copy
6
5 4 5 4 4 5
output
Copy
69
6 1 3 5 2 4
input
Copy
2
1 4
output
Copy
3
2 1
Note

In the first example Vasya can start shooting from the first can. He knocks it down with the first shot because he haven't knocked any other cans down before. After that he has to shoot the third can. To knock it down he shoots 201+1=2120⋅1+1=21 times. After that only second can remains. To knock it down Vasya shoots 102+1=2110⋅2+1=21 times. So the total number of shots is 1+21+21=431+21+21=43.

In the second example the order of shooting does not matter because all cans have the same durability.

Code Examples

#1 Code Example with C++ Programming

Code - C++ Programming

#include <bits/stdc++.h>

using namespace std;

int const N = 1e3 + 10;
int n;
pair<int, int> a[N];

int main() {
#ifndef ONLINE_JUDGE
freopen("in", "r", stdin);
#endif

scanf("%d", &n);
for(int i = 0, tmp; i < n; ++i) {
scanf("%d", &tmp);
a[i].first = tmp;
a[i].second = i;
}
sort(a, a + n);
reverse(a, a + n);

int res = 0;
for(int i = 0; i < n; ++i)
res += (a[i].first * i + 1);

printf("%d\n", res);
for(int i = 0; i < n; ++i) {
if(i != 0) printf(" ");
printf("%d", a[i].second + 1);
}
puts("");

return 0;
}
Copy The Code &

Input

cmd
3
20 10 20

Output

cmd
43
1 3 2