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;
}