[CodeForces-985C]

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
4 2 1
2 2 1 2 3 2 2 3
Output
7
Input
2 1 0
10 10
Output
20
Input
1 2 1
5 2
Output
2
Input
3 2 1
1 2 3 4 5 6
Output
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：一共有m=n*k个木板，组成n个木桶，每个木桶k块板子。木桶的容量取决于（就是等于）最短的板子。并且任意两个木桶的容量差不超过l现在告诉你n,k和每个板子的长度ai。求组成桶的容量和最大是多少？（无法满足条件就输出0）
主要是贪心+模拟；
将最小的板长保存于min。将所有可以作为最低板的木板挑选出来（ai<=min+l）组成一个集合。排序后，在保证每个木桶可以分配到集合里剩下的木板的条件下，尽可能把小的木板分配在一起（一个木桶）。

代码如下

#include
#include
#define max 10000000000
using namespace std;
long long int a[100005];
long long int b[100005];
int cmp(long long int a,long long int b)
{
	return a<b;
}
int main()
{
	long long int i,j,m,n,k,l,min,sum,num;
	scanf("%lld%lld%lld",&n,&k,&l);
	m=n*k;
	sum=0;min=max;
	for(i=1;i<=m;i++)
	{
		scanf("%lld",&a[i]);
		if(min>a[i]) min=a[i];
	}
	j=0;
	for(i=1;i<=m;i++)
	{
		if(a[i]<=min+l)
		{
			j++;
			b[j]=a[i];
		}
	}
	num=j;
	if(num<n) { printf("0"); return 0;}
	sort(b+1,b+num+1,cmp);
	i=1;j=0;
	while(i<=num)
	{
		if(num-i+1-(n-j)+1>=k)
		{
			sum+=b[i];
			j++;i+=k;
		}
		else if(num-i+1>n-j)
		{
			sum+=b[i];
			j++;i+=num-i+1-(n-j);
		}
		else
		{
			sum+=b[i];
			j++;i++;
		}
	}
	printf("%lld",sum);
	return 0;
}