#include <bits/stdc++.h>
using namespace std;
int N, K, L;
int arr[100100];
int main() {
  cin >> N >> K >> L;
  for (int i = 0; i < N * K; i++) cin >> arr[i];
  sort(arr, arr + N * K);
  int cloc = 0;
  while (cloc < N * K && arr[cloc] <= arr[0] + L) cloc++;
  cloc--;
  if (cloc < N - 1)
    cout << "0\n";
  else {
    long long res = 0;
    for (int i = 0; i < N; i++) res += min(arr[i * K], arr[cloc - N + 1 + i]);
    cout << res << "\n";
  }
}

### Prompt

Please formulate a Cpp solution to the following problem:

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.

<image>

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int N, K, L;
int arr[100100];
int main() {
  cin >> N >> K >> L;
  for (int i = 0; i < N * K; i++) cin >> arr[i];
  sort(arr, arr + N * K);
  int cloc = 0;
  while (cloc < N * K && arr[cloc] <= arr[0] + L) cloc++;
  cloc--;
  if (cloc < N - 1)
    cout << "0\n";
  else {
    long long res = 0;
    for (int i = 0; i < N; i++) res += min(arr[i * K], arr[cloc - N + 1 + i]);
    cout << res << "\n";
  }
}

```