coderforce Educational Codeforces Round 44 (Rated for Div. 2) C(赛后补题)

C. Liebig's Barrels

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You have m = n·k wooden staves. The i-th stave has length a_i. You have to assemble n barrels consisting of k staves each, you can use any k staves to construct a barrel. Each stave must belong to exactly one barrel.

Let volume v_j of barrel j be equal to the length of the minimal stave in it.

You want to assemble exactly n barrels with the maximal total sum of volumes. But you have to make them equal enough, so a difference between volumes of any pair of the resulting barrels must not exceed l, i.e. |v_x - v_y| ≤ l for any 1 ≤ x ≤ n and 1 ≤ y ≤ n.

Print maximal total sum of volumes of equal enough barrels or 0 if it's impossible to satisfy the condition above.

Input

The first line contains three space-separated integers n, k and l (1 ≤ n, k ≤ 10⁵, 1 ≤ n·k ≤ 10⁵, 0 ≤ l ≤ 10⁹).

The second line contains m = n·k space-separated integers a₁, a₂, ..., a_m (1 ≤ a_i ≤ 10⁹) — lengths of staves.

Output

Print single integer — maximal total sum of the volumes of barrels or 0 if it's impossible to construct exactly n barrels satisfying the condition |v_x - v_y| ≤ l for any 1 ≤ x ≤ n and 1 ≤ y ≤ n.

Examples

Input

Copy

4 2 1
2 2 1 2 3 2 2 3

Output

Copy

7

Input

Copy

2 1 0
10 10

Output

Copy

20

Input

Copy

1 2 1
5 2

Output

Copy

2

Input

Copy

3 2 1
1 2 3 4 5 6

Output

Copy

0

Note

In the first example you can form the following barrels: [1, 2], [2, 2], [2, 3], [2, 3].

In the second example you can form the following barrels: [10], [10].

In the third example you can form the following barrels: [2, 5].

In the fourth example difference between volumes of barrels in any partition is at least 2 so it is impossible to make barrels equal enough.

这道题忍住了看题解。自己独立思考了QAQ。

但是div2的C题还是属于那种自己可以做但需要花很长时间的那种，希望慢慢进步（D题先凑合着题解自己写QAQ）

补完了这道题去洗澡澡

题意很简单，给n,k,l, l是体积限定的条件见上，k是每个小序列的长度，n是小序列的个数。再给你N*K个数

这道题是一道贪心的题目，我的想法是对于这一系列数字来说：对于分成的k个序列，每个序列中最小的就是代表这个体积，在符合情况| Vi-Vj |<=l的情况下取尽可能大的体积。

首先对数组排序，首先排序后的arr[1]肯定是一个序列的体积（放哪里都是最小的），那么我要满足l的限定条件的话，体积的取值一定在arr[1]~arr[1]+l中，找到arr[1]+l这个值所对应的数组下标。

对于一般情况，n

然后只需要考虑每个序列最小体积地取值，剩余地自然会从剩下的值中取。

再考虑涵盖不掉全部的情况，比如刚才的例子，对于刚刚的6来说，只剩下6和7了，自然涵盖完，但是如果n是3呢，就只能一个取6 一个取7，单独判断一下就好（k不行变成k-1一直减到0为止（保证k过后剩下的元素个数大于等于还没取得序列个数））

说了一大堆，上代码（懒得加注释了改了好多遍才过之前一直WA7是用样例推得时候最小为1,就一直l+1到好久....）

菜啊qwq

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