#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 9, M = 1e9 + 7;
int n, k, l, a[N];
int main() {
  scanf("%d%d%d", &n, &k, &l);
  vector<int> a;
  for (int b, i = 0; i < n * k; i++) {
    scanf("%d", &b);
    a.push_back(b);
  }
  sort(a.begin(), a.end());
  if (n == 1) {
    cout << a[0] << endl;
    return 0;
  }
  int mn = a[0];
  int idx = n * k - k;
  int mx = a[n * k - k];
  while (mx - mn > l) {
    idx--;
    mx = a[idx];
  }
  if (idx < n - 1) return puts("0"), 0;
  stack<int> s;
  long long ans = a[0];
  s.push(idx);
  for (int i = 1; i <= n - 2; i++) {
    s.push(idx - i);
    if (idx - i == 0) return puts("0"), 0;
  }
  int lst = 0;
  while (1) {
    if (s.size() && s.top() - lst > k) {
      int x = s.top();
      s.pop();
      ans += a[lst + k];
      lst += k;
    } else
      break;
  }
  while (s.size()) ans += a[s.top()], s.pop();
  cout << ans << endl;
}

### Prompt

In cpp, your task is to solve 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;
const int N = 1e5 + 9, M = 1e9 + 7;
int n, k, l, a[N];
int main() {
  scanf("%d%d%d", &n, &k, &l);
  vector<int> a;
  for (int b, i = 0; i < n * k; i++) {
    scanf("%d", &b);
    a.push_back(b);
  }
  sort(a.begin(), a.end());
  if (n == 1) {
    cout << a[0] << endl;
    return 0;
  }
  int mn = a[0];
  int idx = n * k - k;
  int mx = a[n * k - k];
  while (mx - mn > l) {
    idx--;
    mx = a[idx];
  }
  if (idx < n - 1) return puts("0"), 0;
  stack<int> s;
  long long ans = a[0];
  s.push(idx);
  for (int i = 1; i <= n - 2; i++) {
    s.push(idx - i);
    if (idx - i == 0) return puts("0"), 0;
  }
  int lst = 0;
  while (1) {
    if (s.size() && s.top() - lst > k) {
      int x = s.top();
      s.pop();
      ans += a[lst + k];
      lst += k;
    } else
      break;
  }
  while (s.size()) ans += a[s.top()], s.pop();
  cout << ans << endl;
}

```