#include <bits/stdc++.h>
using namespace std;
const int _max = 1e5 + 50;
int a[_max];
int main() {
  int n, k, lr;
  while (cin >> n >> k >> lr) {
    int m = n * k;
    for (int i = 1; i <= m; i++) cin >> a[i];
    sort(a + 1, a + m + 1);
    if (a[n] - a[1] > lr) {
      cout << "0" << endl;
      continue;
    }
    long long cnt = a[1] + lr;
    int l = 1, r = m, mid;
    while (l < r) {
      mid = (l + r) >> 1;
      if (cnt < a[mid])
        r = mid;
      else
        l = mid + 1;
    }
    mid = (l + r) >> 1;
    if (cnt < a[mid]) mid--;
    cnt = 0;
    if (k - 1 >= 1) {
      l = (m - mid) / (k - 1);
      for (int i = 0; i < l; i++) {
        cnt += a[mid - i];
      }
      mid -= l;
      l = (m - (mid + l)) % (k - 1);
      l = k - 1 - l;
      mid -= l;
      cnt += a[mid];
      mid--;
    }
    while (mid >= 1) {
      mid -= (k - 1);
      cnt += a[mid--];
    }
    cout << cnt << endl;
  }
  return 0;
}

### Prompt

Your challenge is to write 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;
const int _max = 1e5 + 50;
int a[_max];
int main() {
  int n, k, lr;
  while (cin >> n >> k >> lr) {
    int m = n * k;
    for (int i = 1; i <= m; i++) cin >> a[i];
    sort(a + 1, a + m + 1);
    if (a[n] - a[1] > lr) {
      cout << "0" << endl;
      continue;
    }
    long long cnt = a[1] + lr;
    int l = 1, r = m, mid;
    while (l < r) {
      mid = (l + r) >> 1;
      if (cnt < a[mid])
        r = mid;
      else
        l = mid + 1;
    }
    mid = (l + r) >> 1;
    if (cnt < a[mid]) mid--;
    cnt = 0;
    if (k - 1 >= 1) {
      l = (m - mid) / (k - 1);
      for (int i = 0; i < l; i++) {
        cnt += a[mid - i];
      }
      mid -= l;
      l = (m - (mid + l)) % (k - 1);
      l = k - 1 - l;
      mid -= l;
      cnt += a[mid];
      mid--;
    }
    while (mid >= 1) {
      mid -= (k - 1);
      cnt += a[mid--];
    }
    cout << cnt << endl;
  }
  return 0;
}

```