#include <bits/stdc++.h>
using namespace std;
int M, N, K, L;
const int MAXN = 100010;
int v[MAXN];
int main() {
  scanf("%d%d%d", &N, &K, &L);
  M = N * K;
  for (int i = 0; i < M; ++i) scanf("%d", &v[i]);
  sort(v, v + M);
  int lastgud = upper_bound(v, v + M, v[0] + L) - v - 1;
  if (lastgud + 1 < N) {
    puts("0");
    return 0;
  }
  int barrelsNeeded = N - 1;
  int lastBarrelStart = 0;
  long long ans = v[0];
  while (1) {
    int x = lastgud + 1 - barrelsNeeded;
    if (x - lastBarrelStart <= K) {
      for (int i = x; i <= lastgud; ++i) ans += v[i];
      break;
    }
    int idx = K + lastBarrelStart - 1;
    ++idx;
    lastBarrelStart = idx;
    ans += v[lastBarrelStart];
    --barrelsNeeded;
  }
  printf("%I64d\n", ans);
  return 0;
}

### Prompt

Create a solution in Cpp for 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 M, N, K, L;
const int MAXN = 100010;
int v[MAXN];
int main() {
  scanf("%d%d%d", &N, &K, &L);
  M = N * K;
  for (int i = 0; i < M; ++i) scanf("%d", &v[i]);
  sort(v, v + M);
  int lastgud = upper_bound(v, v + M, v[0] + L) - v - 1;
  if (lastgud + 1 < N) {
    puts("0");
    return 0;
  }
  int barrelsNeeded = N - 1;
  int lastBarrelStart = 0;
  long long ans = v[0];
  while (1) {
    int x = lastgud + 1 - barrelsNeeded;
    if (x - lastBarrelStart <= K) {
      for (int i = x; i <= lastgud; ++i) ans += v[i];
      break;
    }
    int idx = K + lastBarrelStart - 1;
    ++idx;
    lastBarrelStart = idx;
    ans += v[lastBarrelStart];
    --barrelsNeeded;
  }
  printf("%I64d\n", ans);
  return 0;
}

```