Codeforces 985C 贪心

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 ai. 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 vj 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. |vx - vy| ≤ 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, ..., am (1 ≤ ai ≤ 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 |vx - vy| ≤ 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.

题意：木桶容量定义为最短的木板的长度，，用n*k块木板拼成n个木桶，要求每个木桶用k块木板，任意两个木桶容量差距小于l，求能拼成的木桶的最大容积和

思路：贪心，排序之后找到离最短木板差距小于等于l的最长木板，然后尽可能选择长的木板，具体过程见代码。

#include 

using namespace std;

typedef long long ll;
ll a[100005];

int main()
{
    ll n,k,l;
    scanf("%I64d%I64d%I64d",&n,&k,&l);
    for(int i=0; il){ //无法拼出n个桶
        cout<<0<= k-1){    //若仍有大于k-1块木板,则用这k-1块木板拼成桶
            cnt -= k-1;
            pos += k;
        }
        else if(cnt){
            pos += cnt+1;
            cnt = 0;
        }
        else
            pos++;      //每次往后取一块木板
    }
    printf("%I64d\n",res);
    return 0;
}