C. Liebig's Barrels （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

思路：

一道很贪心的贪心题。

比赛时，第一次思路想简单了，我也很感慨我当时能回归正途。

先找出符合 l  的最长的木板，然后木桶装水最多为l，就用比l长的（k-1）块木板，从 l 到 最小 开始组合。如果还有剩余的，然后再从大到小选出k个一组，组合。

代码：

#include <bits/stdc++.h>

using namespace std;
const int maxn=100010;
int a[maxn];

int main()
{
    int n,m,l;
    scanf("%d%d%d",&n,&m,&l);
    long long sum5=0;
    for(int i=1;i<=n*m;i++)
    {
        scanf("%d",&a[i]);
        sum5+=a[i];
    }
    sort(a+1,a+1+n*m);
    if(m==1)
    {
        if(a[n*m]-a[1]<=l)
        {
            printf("%I64d\n",sum5);
        }
        else printf("0\n");
    }
    else
    {
    int st=0;
    priority_queue <int> G;
    for(int i=n*m;i>=n;i--)
    {
        if(a[i]-a[1]<=l)
        {
            st=i;
            break;
        }
        G.push(a[i]);
    }
    if(st==0)
    {
        printf("0\n");
    }
    else
    {
    long long ans=0;
    int su=G.size();
    int ans1=su/(m-1);
    for(int i=st;i>=st-ans1+1;i--)
    {
        int sum=a[i];
        for(int j=1;j<=m-1;j++)
        {
            int sum1=G.top();G.pop();
            sum=min(sum,sum1);
        }
        ans+=sum;
    }
    for(int i=1;i<st-ans1+1;i++)
    {
        G.push(a[i]);
    }
    su=G.size();
    ans1=su/m;
    for(int i=1;i<=ans1;i++)
    {
        int sum1=0x3f3f3f3f;
        for(int j=1;j<=m;j++)
        {
            int sum2=G.top();
            G.pop();
            sum1=min(sum1,sum2);
        }
        ans+=sum1;
    }
    printf("%I64d\n",ans);
    }
    }
    return 0;
}