Technocup 2021 - Elimination Round 2
E
New Game Plus!
Codeforces
2200
原题
← DF →
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Wabbit is playing a game with $n$ bosses numbered from $1$ to $n$. The bosses can be fought in any order. Each boss needs to be defeated exactly once. There is a parameter called boss bonus which is initially $0$.

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When the $i$-th boss is defeated, the current boss bonus is added to Wabbit's score, and then the value of the boss bonus increases by the point increment $c_i$. Note that $c_i$ can be negative, which means that other bosses now give fewer points.

翻译中…

However, Wabbit has found a glitch in the game. At any point in time, he can reset the playthrough and start a New Game Plus playthrough. This will set the current boss bonus to $0$, while all defeated bosses remain defeated. The current score is also saved and does not reset to zero after this operation. This glitch can be used at most $k$ times. He can reset after defeating any number of bosses (including before or after defeating all of them), and he also can reset the game several times in a row without defeating any boss.

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Help Wabbit determine the maximum score he can obtain if he has to defeat all $n$ bosses.

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Input

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The first line of input contains two spaced integers $n$ and $k$ ($1 \leq n \leq 5 \cdot 10^5$, $0 \leq k \leq 5 \cdot 10^5$), representing the number of bosses and the number of resets allowed.

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The next line of input contains $n$ spaced integers $c_1,c_2,\ldots,c_n$ ($-10^6 \leq c_i \leq 10^6$), the point increments of the $n$ bosses.

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Output

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Output a single integer, the maximum score Wabbit can obtain by defeating all $n$ bosses (this value may be negative).

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Note

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In the first test case, no resets are allowed. An optimal sequence of fights would be

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  • Fight the first boss $(+0)$. Boss bonus becomes equal to $1$.
  • Fight the second boss $(+1)$. Boss bonus becomes equal to $2$.
  • Fight the third boss $(+2)$. Boss bonus becomes equal to $3$.
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Thus the answer for the first test case is $0+1+2=3$.

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In the second test case, it can be shown that one possible optimal sequence of fights is

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  • Fight the fifth boss $(+0)$. Boss bonus becomes equal to $5$.
  • Fight the first boss $(+5)$. Boss bonus becomes equal to $4$.
  • Fight the second boss $(+4)$. Boss bonus becomes equal to $2$.
  • Fight the third boss $(+2)$. Boss bonus becomes equal to $-1$.
  • Reset. Boss bonus becomes equal to $0$.
  • Fight the fourth boss $(+0)$. Boss bonus becomes equal to $-4$.
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Hence the answer for the second test case is $0+5+4+2+0=11$.

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样例

Input #1
3 0
1 1 1
Output #1
3
InputCopy OutputCopy
Input #2
5 1
-1 -2 -3 -4 5
Output #2
11
InputCopy OutputCopy
Input #3
13 2
3 1 4 1 5 -9 -2 -6 -5 -3 -5 -8 -9
Output #3
71
InputCopy OutputCopy
题目信息
题号
1415E
时间限制
2 s
内存限制
256 MB
标签
构造
贪心
数学

题面与样例版权归 Codeforces 及出题人所有;中文翻译由 AI 生成(CI 管线),如有出入请以原文为准。