CF573E.Bear and Bowling

Limak is an old brown bear. He often goes bowling with his friends. Today he feels really good and tries to beat his own record!

For rolling a ball one gets a score — an integer (maybe negative) number of points. Score for i-th roll is multiplied by i and scores are summed up. So, for k rolls with scores _s_1, _s_2, ..., s__k, total score is . Total score is 0 if there were no rolls.

Limak made n rolls and got score a__i for i-th of them. He wants to maximize his total score and he came up with an interesting idea. He will cancel some rolls, saying that something distracted him or there was a strong wind.

Limak is able to cancel any number of rolls, maybe even all or none of them. Total score is calculated as if there were only non-canceled rolls. Look at the sample tests for clarification. What maximum total score can Limak get?

Limak 是一只年迈的棕熊。他经常和朋友们一起去打保龄球。今天他感觉特别好，打算打破自己的纪录！

每次投球会得到一个分数——一个整数（可能为负数）分。第 $i$ 次投球的分数将乘以 $i$，然后将所有加权后的分数相加。因此，对于 $k$ 次投球，其分数分别为 $s_1,\,s_2,\,\dots,\,s_k$，总分为
。
若未进行任何投球，则总分为 $0$。

Limak 进行了 $n$ 次投球，第 $i$ 次投球得分为 $a_i$。他希望最大化自己的总分，并想出了一个有趣的方法：他可以取消其中若干次投球，理由可能是“有东西分散了他的注意力”或“当时刮了强风”。

Limak 可以取消任意次数的投球，甚至可以全部取消或一次也不取消。总分仅基于未被取消的投球来计算（即被取消的投球视作从未发生）。请参考样例测试以进一步理解题意。Limak 能获得的最大总分是多少？

The first line contains single integer n (1 ≤ n ≤ 105).

The second line contains n space-separated integers _a_1, _a_2, ..., a__n (|a__i| ≤ 107) - scores for Limak's rolls.

第一行包含一个整数 $n$（$1 \leq n \leq 10^5$）。

第二行包含 $n$ 个以空格分隔的整数 $a_1,\,a_2,\,\ldots,\,a_n$（$|a_i| \leq 10^7$），表示 Limak 每次掷骰子所得的分数。

Print the maximum possible total score after choosing rolls to cancel.

输出选择取消掷骰子后可能获得的最高总分。

In first sample Limak should cancel rolls with scores  - 8 and  - 3. Then he is left with three rolls with scores  - 2, 0, 5. Total score is 1·( - 2) + 2·0 + 3·5 = 13.

In second sample Limak should cancel roll with score  - 50. Total score is 1·( - 10) + 2·20 + 3·( - 30) + 4·40 + 5·60 = 400.

在第一个样例中，Limak 应当取消得分分别为 $-8$ 和 $-3$ 的两次投掷。之后他剩下三次投掷，得分分别为 $-2$、$0$、$5$。总得分为 $1\cdot(-2) + 2\cdot0 + 3\cdot5 = 13$。

在第二个样例中，Limak 应当取消得分为 $-50$ 的一次投掷。总得分为 $1\cdot(-10) + 2\cdot20 + 3\cdot(-30) + 4\cdot40 + 5\cdot60 = 400$。