Liebig's Barrels（CF-ED#42-C）

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：

给你n，k，L

给你n×k个木板要你组成n个木桶，然后每一个木桶的容量差值一定要在L的范围内。

题解：

贪心做法：for（1～n）里面一定是最小的，但是怎么取最优呢，以下是SQY学长的做法：

#include
using namespace std;
typedef long long ll;
const int maxn=1e6+10;
ll a[maxn],n,m,k,len,upper,x,y,sum,cur,i;
int main()
{
    cin>>n>>k>>len;
    m=n*k;
    for(int i=1;i<=m;i++){
        cin>>a[i];
    }
    sort(a+1,a+1+m);

    if(a[n]-a[1]>len){
        return 0*puts("0");
    }

    if(k==1){
        for(i=1;i<=n;i++)
            sum+=a[i];
    }else{
        while (a[upper] - a[1] <= len && upper <= m) upper++;
        upper--;
        x=(upper-n)/(k-1);
        for( i=1,cur=1  ;   i<=x  ;  i++,cur+=k)sum+=a[cur];
        for( i=1;i




#include
using namespace std;
typedef long long ll;
ll m,n,k,len;
int main()
{
        cin>>n>>k>>len;
        m=n*k;
        ll a[m+10];
        for(int i=1;i<=m;i++){
            cin>>a[i];
        }
        sort(a+1,a+1+m);
        if(a[n]-a[1]>len){
            cout<<0<len){
                    pos=i-1;break;
                }
            }
            for(int i=1;i<=pos&&cnt=n&&cnt