#include <bits/stdc++.h>
using namespace std;
int a[100005];
int main() {
  int n, m, i, j, k, s, t, l;
  long long ans;
  scanf("%d%d%d", &n, &k, &l);
  for (i = 1; i <= n * k; i++) scanf("%d", &a[i]);
  sort(a + 1, a + n * k + 1);
  s = 1;
  while (a[s] <= a[1] + l && s <= n * k) s++;
  ans = 0;
  if (s > n * k) {
    for (i = 1; i <= n * k; i += k) ans += (long long)a[i];
    printf("%I64d\n", ans);
    return 0;
  } else {
    t = n;
    for (i = 1; i < s; i += k) {
      ans += (long long)a[i];
      t--;
    }
    i = s - 1;
    while (t > 0) {
      if ((i - 1) % k == 0) i--;
      if (i == 0) break;
      ans += (long long)a[i];
      i--;
      t--;
      if (i == 0) break;
    }
    if (t != 0)
      printf("0\n");
    else
      printf("%I64d\n", ans);
  }
  return 0;
}

### Prompt

Please formulate 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 a[100005];
int main() {
  int n, m, i, j, k, s, t, l;
  long long ans;
  scanf("%d%d%d", &n, &k, &l);
  for (i = 1; i <= n * k; i++) scanf("%d", &a[i]);
  sort(a + 1, a + n * k + 1);
  s = 1;
  while (a[s] <= a[1] + l && s <= n * k) s++;
  ans = 0;
  if (s > n * k) {
    for (i = 1; i <= n * k; i += k) ans += (long long)a[i];
    printf("%I64d\n", ans);
    return 0;
  } else {
    t = n;
    for (i = 1; i < s; i += k) {
      ans += (long long)a[i];
      t--;
    }
    i = s - 1;
    while (t > 0) {
      if ((i - 1) % k == 0) i--;
      if (i == 0) break;
      ans += (long long)a[i];
      i--;
      t--;
      if (i == 0) break;
    }
    if (t != 0)
      printf("0\n");
    else
      printf("%I64d\n", ans);
  }
  return 0;
}

```