#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int n, k, l;
  cin >> n >> k >> l;
  long long int m = n * k;
  long long int a[m];
  for (int i = 0; i < m; i++) {
    cin >> a[i];
  }
  sort(a, a + m);
  long long int it = upper_bound(a, a + m, a[0] + l) - a;
  it--;
  vector<long long int> vis(m + 1, 0);
  long long int cur = 0, j = m - 1, c = 0, ans = 0;
  for (int i = it; i >= 0; i--) {
    if (vis[i]) {
      continue;
    }
    while (j >= 0 && c < k - 1 && a[j] >= a[i]) {
      if (vis[j]) {
        j--;
        continue;
      }
      if (i == j) {
        break;
      }
      vis[j] = 1;
      j--;
      c++;
    }
    if (cur < n && c == k - 1) {
      ans += a[i];
      c = 0;
      vis[i] = 1;
      cur++;
    }
  }
  if (cur == n) {
    cout << ans << "\n";
  } else {
    cout << "0\n";
  }
  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() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int n, k, l;
  cin >> n >> k >> l;
  long long int m = n * k;
  long long int a[m];
  for (int i = 0; i < m; i++) {
    cin >> a[i];
  }
  sort(a, a + m);
  long long int it = upper_bound(a, a + m, a[0] + l) - a;
  it--;
  vector<long long int> vis(m + 1, 0);
  long long int cur = 0, j = m - 1, c = 0, ans = 0;
  for (int i = it; i >= 0; i--) {
    if (vis[i]) {
      continue;
    }
    while (j >= 0 && c < k - 1 && a[j] >= a[i]) {
      if (vis[j]) {
        j--;
        continue;
      }
      if (i == j) {
        break;
      }
      vis[j] = 1;
      j--;
      c++;
    }
    if (cur < n && c == k - 1) {
      ans += a[i];
      c = 0;
      vis[i] = 1;
      cur++;
    }
  }
  if (cur == n) {
    cout << ans << "\n";
  } else {
    cout << "0\n";
  }
  return 0;
}

```