#include <bits/stdc++.h>
using namespace std;
void solve(int testcase) {
  long long int n, k, L;
  cin >> n >> k >> L;
  long long int m = n * k;
  vector<long long int> a(m);
  for (int i = (0); i <= (m - 1); i++) cin >> a[i];
  sort(a.begin(), a.end());
  int id = -1;
  long long int what = a[0] + L;
  for (int i = (0); i <= (m - 1); i++) {
    if (a[i] <= what) id++;
  }
  if (id + 1 < n) {
    cout << '0';
    return;
  }
  long long int sum = 0, done = n, rem = m - id - 1;
  int i = id;
  while (done && i >= 0) {
    if (rem >= k - 1 && i >= done - 1) {
      sum += a[i];
      i--;
      done--;
      rem -= (k - 1);
    } else {
      rem++;
      i--;
    }
  }
  cout << sum;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cout << setprecision(10);
  clock_t b = clock();
  int t = 1;
  for (int tt = (1); tt <= (t); tt++) solve(tt);
  clock_t e = clock();
  cerr << (double(e - b) / CLOCKS_PER_SEC) << " sec";
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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;
void solve(int testcase) {
  long long int n, k, L;
  cin >> n >> k >> L;
  long long int m = n * k;
  vector<long long int> a(m);
  for (int i = (0); i <= (m - 1); i++) cin >> a[i];
  sort(a.begin(), a.end());
  int id = -1;
  long long int what = a[0] + L;
  for (int i = (0); i <= (m - 1); i++) {
    if (a[i] <= what) id++;
  }
  if (id + 1 < n) {
    cout << '0';
    return;
  }
  long long int sum = 0, done = n, rem = m - id - 1;
  int i = id;
  while (done && i >= 0) {
    if (rem >= k - 1 && i >= done - 1) {
      sum += a[i];
      i--;
      done--;
      rem -= (k - 1);
    } else {
      rem++;
      i--;
    }
  }
  cout << sum;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cout << setprecision(10);
  clock_t b = clock();
  int t = 1;
  for (int tt = (1); tt <= (t); tt++) solve(tt);
  clock_t e = clock();
  cerr << (double(e - b) / CLOCKS_PER_SEC) << " sec";
  return 0;
}

```