#include <bits/stdc++.h>
using namespace std;
const int N = (int)1e9;
int main(int argc, char *argv[]) {
  int n, k, l;
  scanf("%d %d %d", &n, &k, &l);
  int m = n * k;
  int a[m];
  for (int i = 0; i < m; ++i) {
    scanf("%d", &a[i]);
  }
  sort(a, a + m);
  long long ans = a[0];
  for (int i = m - 1; i > 0; --i) {
    if (a[i] - a[0] <= l) {
      if (i < n - 1) {
        ans = 0;
        break;
      }
      int x = m - i;
      while (n > 1) {
        if (x >= k) {
          ans += a[i];
          x -= k;
          --n;
        }
        ++x;
        --i;
      }
      break;
    }
  }
  printf("%I64d\n", ans);
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;
const int N = (int)1e9;
int main(int argc, char *argv[]) {
  int n, k, l;
  scanf("%d %d %d", &n, &k, &l);
  int m = n * k;
  int a[m];
  for (int i = 0; i < m; ++i) {
    scanf("%d", &a[i]);
  }
  sort(a, a + m);
  long long ans = a[0];
  for (int i = m - 1; i > 0; --i) {
    if (a[i] - a[0] <= l) {
      if (i < n - 1) {
        ans = 0;
        break;
      }
      int x = m - i;
      while (n > 1) {
        if (x >= k) {
          ans += a[i];
          x -= k;
          --n;
        }
        ++x;
        --i;
      }
      break;
    }
  }
  printf("%I64d\n", ans);
  return 0;
}

```