#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, k, l;
  cin >> n >> k >> l;
  map<long long, int> s;
  for (int i = 0; i < n * k; i++) {
    long long c;
    cin >> c;
    if (s.find(c) != s.end())
      s[c]++;
    else
      s.insert(make_pair(c, 1));
  }
  long long first = -1;
  int num = 0;
  deque<long long> dq;
  for (map<long long, int>::iterator it = s.begin(); it != s.end(); it++) {
    if (first < 0) first = it->first;
    if (it->first - first > l) break;
    num += it->second;
    for (int i = 0; i < it->second; i++) {
      dq.push_back(it->first);
    }
  }
  if (num < n) {
    cout << 0 << endl;
    return 0;
  }
  long long ans = 0;
  int t = 0;
  for (int i = 0; i < dq.size() && t < n;) {
    ans += dq[i];
    t++;
    if (num - n > 0) {
      i += min(k - 1, num - n) + 1;
      num -= min(k - 1, num - n);
    } else
      i++;
  }
  cout << ans << endl;
  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;
int main() {
  int n, k, l;
  cin >> n >> k >> l;
  map<long long, int> s;
  for (int i = 0; i < n * k; i++) {
    long long c;
    cin >> c;
    if (s.find(c) != s.end())
      s[c]++;
    else
      s.insert(make_pair(c, 1));
  }
  long long first = -1;
  int num = 0;
  deque<long long> dq;
  for (map<long long, int>::iterator it = s.begin(); it != s.end(); it++) {
    if (first < 0) first = it->first;
    if (it->first - first > l) break;
    num += it->second;
    for (int i = 0; i < it->second; i++) {
      dq.push_back(it->first);
    }
  }
  if (num < n) {
    cout << 0 << endl;
    return 0;
  }
  long long ans = 0;
  int t = 0;
  for (int i = 0; i < dq.size() && t < n;) {
    ans += dq[i];
    t++;
    if (num - n > 0) {
      i += min(k - 1, num - n) + 1;
      num -= min(k - 1, num - n);
    } else
      i++;
  }
  cout << ans << endl;
  return 0;
}

```