#include <bits/stdc++.h>
using namespace std;
int main() {
  long int n, k;
  long long int d, sum = 0;
  cin >> n >> k >> d;
  long long int m = n * k;
  long long int arr[100001];
  for (long int i = 0; i < (m); i++) cin >> arr[i];
  sort(arr, arr + (m));
  if (arr[n - 1] - arr[0] > d)
    cout << 0;
  else {
    long long int sum1 = 0;
    vector<long long int> vec;
    for (int i = 0; i < m; i++) {
      if (arr[i] - arr[0] <= d) {
        vec.push_back(arr[i]);
        sum1 += arr[i];
      } else
        break;
    }
    if (vec.size() == n)
      cout << sum1;
    else {
      long long int ans = 0;
      long long int n1 = m - vec.size();
      long long int n2 = n1 % (k - 1);
      long long int n3 = n1 / (k - 1);
      long int ctr = 0, index;
      for (long int i = vec.size() - 1; i >= 0; i--) {
        if (ctr < n3) {
          ans += vec[i];
          ctr++;
        } else {
          index = i;
          break;
        }
      }
      long long int stop;
      if (n2 != 0) {
        stop = index + 1 - (k - n2);
        ans += vec[index - (k - n2) + 1];
      } else
        stop = index;
      for (long int i = 0; i < stop; i = i + k) ans += vec[i];
      cout << ans;
    }
  }
}

### Prompt

Please formulate 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 int n, k;
  long long int d, sum = 0;
  cin >> n >> k >> d;
  long long int m = n * k;
  long long int arr[100001];
  for (long int i = 0; i < (m); i++) cin >> arr[i];
  sort(arr, arr + (m));
  if (arr[n - 1] - arr[0] > d)
    cout << 0;
  else {
    long long int sum1 = 0;
    vector<long long int> vec;
    for (int i = 0; i < m; i++) {
      if (arr[i] - arr[0] <= d) {
        vec.push_back(arr[i]);
        sum1 += arr[i];
      } else
        break;
    }
    if (vec.size() == n)
      cout << sum1;
    else {
      long long int ans = 0;
      long long int n1 = m - vec.size();
      long long int n2 = n1 % (k - 1);
      long long int n3 = n1 / (k - 1);
      long int ctr = 0, index;
      for (long int i = vec.size() - 1; i >= 0; i--) {
        if (ctr < n3) {
          ans += vec[i];
          ctr++;
        } else {
          index = i;
          break;
        }
      }
      long long int stop;
      if (n2 != 0) {
        stop = index + 1 - (k - n2);
        ans += vec[index - (k - n2) + 1];
      } else
        stop = index;
      for (long int i = 0; i < stop; i = i + k) ans += vec[i];
      cout << ans;
    }
  }
}

```