#include <bits/stdc++.h>
using namespace std;
inline bool equal(double _x, double _y) {
  return _x > _y - 1e-9 && _x < _y + 1e-9;
}
int n, k, l;
long long a, sum, b;
multiset<long long> s;
int main() {
  scanf("%d%d%d", &n, &k, &l);
  for (int i = 0; i < n * k; i++) {
    scanf("%lld", &a);
    s.insert(a);
  }
  b = *s.begin();
  if (n == 1) {
    printf("%lld\n", *s.begin());
    return 0;
  }
  bool f = 0;
  while (!s.empty()) {
    auto tt = s.end();
    tt--;
    for (int i = 0; i < k - 1; i++) {
      auto tp = tt;
      if (tp != s.begin()) tp--;
      s.erase(tt);
      tt = tp;
    }
    auto it = s.upper_bound(b + l);
    if (it == s.begin()) {
      f = 1;
      break;
    }
    it--;
    sum += *it;
    s.erase(it);
  }
  if (f == 0)
    printf("%lld\n", sum);
  else
    puts("0");
}

### 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 bool equal(double _x, double _y) {
  return _x > _y - 1e-9 && _x < _y + 1e-9;
}
int n, k, l;
long long a, sum, b;
multiset<long long> s;
int main() {
  scanf("%d%d%d", &n, &k, &l);
  for (int i = 0; i < n * k; i++) {
    scanf("%lld", &a);
    s.insert(a);
  }
  b = *s.begin();
  if (n == 1) {
    printf("%lld\n", *s.begin());
    return 0;
  }
  bool f = 0;
  while (!s.empty()) {
    auto tt = s.end();
    tt--;
    for (int i = 0; i < k - 1; i++) {
      auto tp = tt;
      if (tp != s.begin()) tp--;
      s.erase(tt);
      tt = tp;
    }
    auto it = s.upper_bound(b + l);
    if (it == s.begin()) {
      f = 1;
      break;
    }
    it--;
    sum += *it;
    s.erase(it);
  }
  if (f == 0)
    printf("%lld\n", sum);
  else
    puts("0");
}

```