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.

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.

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.

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.

思路：

　　先找把每k个中最小的加起来，这是必须的。然后再把符合要求的从上至下，加在一起。

　　贪心思想。

　　不懂请留言，没人看不想写得太详细，况且这题本身不难。

#include<iostream>

#include<algorithm>

using namespace std;

long long a[];

bool book[];

int main()

{

    int n,k,l;

    cin>>n>>k>>l;

    for(int i=;i<=n*k;i++){

        cin>>a[i];

    }

    sort(a+,a+n*k+);

    int flag=;

    int maxx,minn,t=n*k;

    long long ans=;

    if(a[n]-a[]>l){cout<<<<endl;}

    else{

        for(int i=n;i<=n*k;i++){

            if(a[i]-a[]>l){t=i-;break;}

        }

        int flag=;

        for(int j=;j<=t;j+=k){

            ans+=a[j];flag++;

            book[j]=true;

            if(flag==n){break;}

        }

        for(int j=t;j>=;j--){

            if(flag==n){break;}

            if(!book[j]){ans+=a[j];flag++;}

            if(flag==n){break;}

        }

        cout<<ans<<endl;

    }

}