C. Liebig's Barrels

You have m = n·k wooden staves. The i-th stave has length a_i. 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 v_j 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. |v_x - v_y| ≤ 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, ..., a_m (1 ≤ a_i ≤ 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 |v_x - v_y| ≤ 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.

#include 
#include
#include
#include
#include
using namespace std;
typedef long long ll;

const int maxn=1e5+10;
ll val[maxn];

/*
挺好的一道思维题和模拟题。。。
*/

int main()
{
    ll n,m,l;
    scanf("%lld %lld %lld",&n,&m,&l);
    for(int i=1;i<=n*m;i++){
       scanf("%lld",&val[i]);
    }
    sort(val+1,val+n*m+1);
    int pos=lower_bound(val+1,val+n*m+1,val[1]+l+1)-val;
    pos--;

    //for(int i=1;i<=n*m;i++)printf("i:%d %lld\n",i,val[i]);
    //printf("pos:%d\n",pos);

    ll sum=0;
    if(pos>=n*m){
        for(int i=n*m-m+1;i>=1;i-=m)
            sum+=val[i];
    }
    else{
        int res=n*m-pos,cur=0;
        for(int i=pos;i>=1;){
            cur=0;
            while(cur