#include <bits/stdc++.h>
const long double eps = 0.00000001;
const long long MOD = 1e9 + 7;
using namespace std;
long long ans;
int main() {
  fflush(stdin);
  cout << fixed, cout.precision(18);
  ios_base::sync_with_stdio(false);
  long long i, j, n, k, m, l;
  string second;
  cin >> n >> k >> l;
  m = n * k;
  long long a[m];
  multiset<long long> se;
  for (i = 0; i < m; ++i) cin >> a[i], se.insert(a[i]);
  sort(a, a + m);
  for (i = m - k; i >= 0; i -= k) {
    auto it = se.upper_bound(a[0] + l);
    if (it != se.begin()) {
      --it;
      int best = *it;
      se.erase(se.find(*it));
      if (a[i] > best) a[i] = best;
      ans += a[i];
    } else
      return cout << 0, 0;
  }
  cout << ans;
}

### Prompt

In cpp, your task is to solve 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>
const long double eps = 0.00000001;
const long long MOD = 1e9 + 7;
using namespace std;
long long ans;
int main() {
  fflush(stdin);
  cout << fixed, cout.precision(18);
  ios_base::sync_with_stdio(false);
  long long i, j, n, k, m, l;
  string second;
  cin >> n >> k >> l;
  m = n * k;
  long long a[m];
  multiset<long long> se;
  for (i = 0; i < m; ++i) cin >> a[i], se.insert(a[i]);
  sort(a, a + m);
  for (i = m - k; i >= 0; i -= k) {
    auto it = se.upper_bound(a[0] + l);
    if (it != se.begin()) {
      --it;
      int best = *it;
      se.erase(se.find(*it));
      if (a[i] > best) a[i] = best;
      ans += a[i];
    } else
      return cout << 0, 0;
  }
  cout << ans;
}

```