#include <bits/stdc++.h>
using namespace std;
inline int read() {
  int x;
  char c;
  while ((c = getchar()) < '0' || c > '9')
    ;
  for (x = c - '0'; (c = getchar()) >= '0' && c <= '9';) x = x * 10 + c - '0';
  return x;
}
int a[100000 + 5];
int main() {
  int n, k, l, i, x;
  long long ans = 0;
  n = read();
  k = read();
  l = read();
  for (i = 1; i <= n * k; ++i) a[i] = read();
  sort(a + 1, a + n * k + 1);
  x = upper_bound(a + 1, a + n * k + 1, a[1] + l) - a - 1;
  if (x < n) return 0 * puts("0");
  for (i = 1; i <= n; ++i) ans += a[min(x - i + 1, (n - i) * k + 1)];
  cout << ans;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
inline int read() {
  int x;
  char c;
  while ((c = getchar()) < '0' || c > '9')
    ;
  for (x = c - '0'; (c = getchar()) >= '0' && c <= '9';) x = x * 10 + c - '0';
  return x;
}
int a[100000 + 5];
int main() {
  int n, k, l, i, x;
  long long ans = 0;
  n = read();
  k = read();
  l = read();
  for (i = 1; i <= n * k; ++i) a[i] = read();
  sort(a + 1, a + n * k + 1);
  x = upper_bound(a + 1, a + n * k + 1, a[1] + l) - a - 1;
  if (x < n) return 0 * puts("0");
  for (i = 1; i <= n; ++i) ans += a[min(x - i + 1, (n - i) * k + 1)];
  cout << ans;
}

```