#include <bits/stdc++.h>
int n, k, m;
long long l;
long long a[100010];
bool vis[100010];
using namespace std;
long long solve() {
  if (a[n] - a[1] > l) return 0;
  long long tot = 0, mark, left;
  left = n;
  for (int i = 1; i <= m; i++)
    if (a[i] - a[1] > l) {
      mark = i - 1;
      break;
    }
  for (int i = 1; i <= mark; i += k) {
    tot += a[i];
    left--;
    vis[i] = true;
    if (!left) return tot;
  }
  for (int i = mark; i >= 1; i--)
    if (!vis[i]) {
      tot += a[i];
      left--;
      if (!left) return tot;
    }
}
int main() {
  scanf("%d%d%lld", &n, &k, &l);
  m = n * k;
  for (int i = 1; i <= m; i++) scanf("%lld", &a[i]);
  sort(a + 1, a + 1 + m);
  printf("%lld", solve());
}

### Prompt

Your task is to create 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>
int n, k, m;
long long l;
long long a[100010];
bool vis[100010];
using namespace std;
long long solve() {
  if (a[n] - a[1] > l) return 0;
  long long tot = 0, mark, left;
  left = n;
  for (int i = 1; i <= m; i++)
    if (a[i] - a[1] > l) {
      mark = i - 1;
      break;
    }
  for (int i = 1; i <= mark; i += k) {
    tot += a[i];
    left--;
    vis[i] = true;
    if (!left) return tot;
  }
  for (int i = mark; i >= 1; i--)
    if (!vis[i]) {
      tot += a[i];
      left--;
      if (!left) return tot;
    }
}
int main() {
  scanf("%d%d%lld", &n, &k, &l);
  m = n * k;
  for (int i = 1; i <= m; i++) scanf("%lld", &a[i]);
  sort(a + 1, a + 1 + m);
  printf("%lld", solve());
}

```