#include <bits/stdc++.h>
int n, k, l, m, tot, a[100005];
bool vis[100005];
long long ans;
int main() {
  scanf("%d%d%d", &n, &k, &l);
  m = n * k;
  for (int i = 1; i <= m; ++i) scanf("%d", &a[i]);
  std::sort(a + 1, a + m + 1);
  for (int i = 1; i <= m && a[i] - a[1] <= l; i += k)
    tot++, vis[i] = 1, ans += a[i];
  for (int i = m; i && tot < n; --i)
    if (a[i] - a[1] <= l && !vis[i]) tot++, vis[i] = 1, ans += a[i];
  printf("%I64d\n", tot == n ? ans : 0ll);
}

### 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, l, m, tot, a[100005];
bool vis[100005];
long long ans;
int main() {
  scanf("%d%d%d", &n, &k, &l);
  m = n * k;
  for (int i = 1; i <= m; ++i) scanf("%d", &a[i]);
  std::sort(a + 1, a + m + 1);
  for (int i = 1; i <= m && a[i] - a[1] <= l; i += k)
    tot++, vis[i] = 1, ans += a[i];
  for (int i = m; i && tot < n; --i)
    if (a[i] - a[1] <= l && !vis[i]) tot++, vis[i] = 1, ans += a[i];
  printf("%I64d\n", tot == n ? ans : 0ll);
}

```