#include <bits/stdc++.h>
using namespace std;
class Solution {
 public:
  void solve(std::istream &in, std::ostream &out) {
    long long n, k, l;
    in >> n >> k >> l;
    std::vector<long long> a(n * k);
    for (auto &el : a) in >> el;
    std::sort(a.begin(), a.end());
    std::vector<bool> b(n * k);
    auto l_ind = std::upper_bound(a.begin(), a.end(), a[0] + l) - a.begin();
    long long cnt = 0;
    for (long long i = 0; i < l_ind; i += k) {
      b[i] = true;
      ++cnt;
    }
    if (cnt < n)
      for (long long j = l_ind - 1; j >= 0; j--) {
        if (!b[j]) {
          b[j] = true;
          ++cnt;
        }
        if (cnt == n) break;
      }
    if (cnt < n) return void(out << "0\n");
    long long res = 0;
    for (long long i = 0; i < n * k; ++i)
      if (b[i]) res += a[i];
    out << res << '\n';
  }
};
void solve(std::istream &in, std::ostream &out) {
  out << std::setprecision(12);
  Solution solution;
  solution.solve(in, out);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  istream &in = cin;
  ostream &out = cout;
  solve(in, out);
  return 0;
}

### 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;
class Solution {
 public:
  void solve(std::istream &in, std::ostream &out) {
    long long n, k, l;
    in >> n >> k >> l;
    std::vector<long long> a(n * k);
    for (auto &el : a) in >> el;
    std::sort(a.begin(), a.end());
    std::vector<bool> b(n * k);
    auto l_ind = std::upper_bound(a.begin(), a.end(), a[0] + l) - a.begin();
    long long cnt = 0;
    for (long long i = 0; i < l_ind; i += k) {
      b[i] = true;
      ++cnt;
    }
    if (cnt < n)
      for (long long j = l_ind - 1; j >= 0; j--) {
        if (!b[j]) {
          b[j] = true;
          ++cnt;
        }
        if (cnt == n) break;
      }
    if (cnt < n) return void(out << "0\n");
    long long res = 0;
    for (long long i = 0; i < n * k; ++i)
      if (b[i]) res += a[i];
    out << res << '\n';
  }
};
void solve(std::istream &in, std::ostream &out) {
  out << std::setprecision(12);
  Solution solution;
  solution.solve(in, out);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  istream &in = cin;
  ostream &out = cout;
  solve(in, out);
  return 0;
}

```