Problem:
Score : 700 points

Problem Statement

We have N gemstones labeled 1 through N.

You can perform the following operation any number of times (possibly zero).

Select a positive integer x, and smash all the gems labeled with multiples of x.

Then, for each i, if the gem labeled i remains without getting smashed, you will receive a_i yen (the currency of Japan).
However, a_i may be negative, in which case you will be charged money.

By optimally performing the operation, how much yen can you earn?

Constraints

All input values are integers.

1 \leq N \leq 100

|a_i| \leq 10^9

Input

Input is given from Standard Input in the following format:

N
a_1 a_2 ... a_N

Output

Print the maximum amount of money that can be earned.

Sample Input 1

6
1 2 -6 4 5 3

Sample Output 1

12

It is optimal to smash Gem 3 and 6.

Sample Input 2

6
100 -100 -100 -100 100 -100

Sample Output 2

200

Sample Input 3

5
-1 -2 -3 -4 -5

Sample Output 3

0

It is optimal to smash all the gems.

Sample Input 4

2
-1000 100000

Sample Output 4

99000

Solution in C++14 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include"bits/stdc++.h"
using namespace std;
using ll = int64_t;
ll N;
vector<ll> a;

ll solve(ll i, vector<ll> erases) {
    if (i == N + 1) {
        auto b = a;
        for (ll e : erases) {
            for (ll j = e; j <= N; j += e) {
                b[j] = 0;
            }
        }
        return accumulate(b.begin(), b.end(), (ll)0);
    }

    ll result = LLONG_MIN;

    //erasesの中にすでにiの約数が入っていれば既に叩き割られている
    for (ll e : erases) {
        if (i % e == 0) {
            return solve(i + 1, erases);
        }
    }

    //叩き割らない場合
    result = max(result, solve(i + 1, erases));
    
    //叩き割る場合
    erases.push_back(i);
    result = max(result, solve(i + 1, erases));

    return result;
}

int main() {
    cin >> N;
    a.resize(N + 1, 0);
    for (ll i = 1; i <= N; i++) {
        cin >> a[i];
    }

    cout << solve(1, vector<ll>()) << endl;
}