#include <bits/stdc++.h>
using namespace std;
int main() {
  int N, K, L;
  cin >> N >> K >> L;
  int M = N * K;
  vector<int> all;
  for (int i = 0; i < M; i++) {
    int a;
    cin >> a;
    all.push_back(a);
  }
  sort(all.begin(), all.end());
  int minD = all[N - 1] - all[0];
  if (minD > L) {
    cout << 0;
    return 0;
  }
  vector<int>::iterator it = upper_bound(all.begin(), all.end(), all[0] + L);
  int count = it - all.begin();
  vector<vector<int>> barrels(N);
  vector<int> mi(N, INT_MAX);
  int index = 0;
  int otherside = 0;
  int from = count - 1;
  while (index < N && otherside <= from) {
    while (barrels[index].size() < K && otherside <= from) {
      barrels[index].push_back(all[otherside]);
      if (mi[index] > all[otherside]) mi[index] = all[otherside];
      otherside++;
      if (barrels[index].size() < K && N - index >= from - otherside + 1) break;
    }
    if (barrels[index].size() < K && N - index >= from - otherside + 1) break;
    index++;
  }
  index = N - 1;
  for (int i = from; i >= otherside; i--) {
    barrels[index].push_back(all[i]);
    if (mi[index] > all[i]) mi[index] = all[i];
    index--;
  }
  unsigned long long int sizesum = 0;
  for (int i = 0; i < N; i++) sizesum += mi[i];
  cout << sizesum;
  return 0;
}

### Prompt

Your challenge is to write 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 main() {
  int N, K, L;
  cin >> N >> K >> L;
  int M = N * K;
  vector<int> all;
  for (int i = 0; i < M; i++) {
    int a;
    cin >> a;
    all.push_back(a);
  }
  sort(all.begin(), all.end());
  int minD = all[N - 1] - all[0];
  if (minD > L) {
    cout << 0;
    return 0;
  }
  vector<int>::iterator it = upper_bound(all.begin(), all.end(), all[0] + L);
  int count = it - all.begin();
  vector<vector<int>> barrels(N);
  vector<int> mi(N, INT_MAX);
  int index = 0;
  int otherside = 0;
  int from = count - 1;
  while (index < N && otherside <= from) {
    while (barrels[index].size() < K && otherside <= from) {
      barrels[index].push_back(all[otherside]);
      if (mi[index] > all[otherside]) mi[index] = all[otherside];
      otherside++;
      if (barrels[index].size() < K && N - index >= from - otherside + 1) break;
    }
    if (barrels[index].size() < K && N - index >= from - otherside + 1) break;
    index++;
  }
  index = N - 1;
  for (int i = from; i >= otherside; i--) {
    barrels[index].push_back(all[i]);
    if (mi[index] > all[i]) mi[index] = all[i];
    index--;
  }
  unsigned long long int sizesum = 0;
  for (int i = 0; i < N; i++) sizesum += mi[i];
  cout << sizesum;
  return 0;
}

```