#include <bits/stdc++.h>
using namespace std;
int main() {
  long long n, m, l;
  cin >> n >> m >> l;
  int box = n;
  long long t = n * m, s = 0;
  long long a[5 + t];
  for (int i = 1; i <= n * m; ++i) cin >> a[i];
  sort(a + 1, 1 + a + t);
  if (n == 1) cout << a[1] << endl, exit(0);
  long long b[t + 5];
  b[1] = 0;
  for (int i = 2; i <= t; ++i) {
    b[i] = (b[i - 1] + a[i] - a[i - 1]);
  }
  long long last = t;
  for (int i = 1; i <= t; ++i) {
    if (b[i] > l) {
      last = i - 1;
      break;
    }
  }
  if (last < n)
    cout << 0 << endl;
  else if (last == n) {
    long long sum = 0;
    while (last >= 1) {
      sum += a[last--];
    }
    cout << sum << endl;
  } else {
    long long beshi = (last - n);
    long long per_box1 = beshi / m;
    long long per_box2 = per_box1;
    long long extra = beshi % m, ans = 0;
    long long Box = 1;
    for (int i = 1; Box <= last; ++i) {
      ans += a[Box++];
      for (int j = 1; j < m; ++j) {
        if (last - Box > n - i - 1)
          Box++;
        else
          break;
      }
    }
    cout << ans << endl;
  }
}

### Prompt

Generate 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() {
  long long n, m, l;
  cin >> n >> m >> l;
  int box = n;
  long long t = n * m, s = 0;
  long long a[5 + t];
  for (int i = 1; i <= n * m; ++i) cin >> a[i];
  sort(a + 1, 1 + a + t);
  if (n == 1) cout << a[1] << endl, exit(0);
  long long b[t + 5];
  b[1] = 0;
  for (int i = 2; i <= t; ++i) {
    b[i] = (b[i - 1] + a[i] - a[i - 1]);
  }
  long long last = t;
  for (int i = 1; i <= t; ++i) {
    if (b[i] > l) {
      last = i - 1;
      break;
    }
  }
  if (last < n)
    cout << 0 << endl;
  else if (last == n) {
    long long sum = 0;
    while (last >= 1) {
      sum += a[last--];
    }
    cout << sum << endl;
  } else {
    long long beshi = (last - n);
    long long per_box1 = beshi / m;
    long long per_box2 = per_box1;
    long long extra = beshi % m, ans = 0;
    long long Box = 1;
    for (int i = 1; Box <= last; ++i) {
      ans += a[Box++];
      for (int j = 1; j < m; ++j) {
        if (last - Box > n - i - 1)
          Box++;
        else
          break;
      }
    }
    cout << ans << endl;
  }
}

```