#include <bits/stdc++.h>
using namespace std;
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long n, k, l;
  cin >> n >> k >> l;
  vector<long long> tab(n * k);
  for (long long i = 0; i < n * k; i++) {
    cin >> tab[i];
  }
  sort(tab.begin(), tab.end());
  vector<long long> liste(0);
  long long mini = tab[0] + l;
  for (long long i = 0; i < n * k; i++) {
    if (tab[i] > mini) break;
    liste.push_back(tab[i]);
  }
  if ((k - 1) * liste.size() < tab.size() - liste.size()) {
    cout << 0 << endl;
    return 0;
  }
  long long tot = 0;
  long long deb = liste.size();
  while (1) {
    if (tab.size() - deb <= k - 1) {
      break;
    }
    for (long long i = 0; i < k - 1; i++) {
      tab.pop_back();
    }
    tot += liste[liste.size() - 1];
    liste.pop_back();
  }
  long long rest = k - (tab.size() - deb);
  while (liste.size() > 0) {
    for (long long i = 0; i < rest - 1; i++) {
      liste.pop_back();
    }
    tot += liste[liste.size() - 1];
    liste.pop_back();
    rest = k;
  }
  cout << tot << endl;
}

### Prompt

Create a solution in Cpp for 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;
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long n, k, l;
  cin >> n >> k >> l;
  vector<long long> tab(n * k);
  for (long long i = 0; i < n * k; i++) {
    cin >> tab[i];
  }
  sort(tab.begin(), tab.end());
  vector<long long> liste(0);
  long long mini = tab[0] + l;
  for (long long i = 0; i < n * k; i++) {
    if (tab[i] > mini) break;
    liste.push_back(tab[i]);
  }
  if ((k - 1) * liste.size() < tab.size() - liste.size()) {
    cout << 0 << endl;
    return 0;
  }
  long long tot = 0;
  long long deb = liste.size();
  while (1) {
    if (tab.size() - deb <= k - 1) {
      break;
    }
    for (long long i = 0; i < k - 1; i++) {
      tab.pop_back();
    }
    tot += liste[liste.size() - 1];
    liste.pop_back();
  }
  long long rest = k - (tab.size() - deb);
  while (liste.size() > 0) {
    for (long long i = 0; i < rest - 1; i++) {
      liste.pop_back();
    }
    tot += liste[liste.size() - 1];
    liste.pop_back();
    rest = k;
  }
  cout << tot << endl;
}

```