#include <bits/stdc++.h>
#pragma GCC optimize(3)
using namespace std;
void qr(int &num) {
  num = 0;
  int y = 1;
  char c;
  c = getchar();
  while ((c < '0' || c > '9') && c != '-') c = getchar();
  if (c == '-') y = -1, c = getchar();
  while (c >= '0' && c <= '9') num = num * 10 + c - '0', c = getchar();
  num *= y;
}
int n, k, l, a[100005];
int main() {
  qr(n);
  qr(k);
  qr(l);
  for (int i = 1; i <= n * k; i++) {
    qr(a[i]);
  }
  sort(a + 1, a + 1 + n * k);
  bool flag = true;
  for (int i = 2; i <= n; i++) {
    if (a[i] - a[1] > l) {
      flag = false;
      break;
    }
  }
  if (!flag) {
    cout << 0;
    return 0;
  }
  int f = 0;
  for (int i = n * k; i >= 1; i--) {
    if (a[i] - a[1] <= l) {
      f = i;
      break;
    }
  }
  int now = 1, tail = n * k, m = f;
  long long cnt = 0;
  for (int i = n * k; i >= 1; i -= (k - 1)) {
    if (i - (k - 1) <= f) break;
    cnt = 1ll * a[m] + cnt;
    m--;
  }
  for (int i = 1; i <= m; i += k) {
    cnt = 1ll * a[i] + cnt;
  }
  cout << cnt;
  return 0;
}

### Prompt

Create a solution in cpp for 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>
#pragma GCC optimize(3)
using namespace std;
void qr(int &num) {
  num = 0;
  int y = 1;
  char c;
  c = getchar();
  while ((c < '0' || c > '9') && c != '-') c = getchar();
  if (c == '-') y = -1, c = getchar();
  while (c >= '0' && c <= '9') num = num * 10 + c - '0', c = getchar();
  num *= y;
}
int n, k, l, a[100005];
int main() {
  qr(n);
  qr(k);
  qr(l);
  for (int i = 1; i <= n * k; i++) {
    qr(a[i]);
  }
  sort(a + 1, a + 1 + n * k);
  bool flag = true;
  for (int i = 2; i <= n; i++) {
    if (a[i] - a[1] > l) {
      flag = false;
      break;
    }
  }
  if (!flag) {
    cout << 0;
    return 0;
  }
  int f = 0;
  for (int i = n * k; i >= 1; i--) {
    if (a[i] - a[1] <= l) {
      f = i;
      break;
    }
  }
  int now = 1, tail = n * k, m = f;
  long long cnt = 0;
  for (int i = n * k; i >= 1; i -= (k - 1)) {
    if (i - (k - 1) <= f) break;
    cnt = 1ll * a[m] + cnt;
    m--;
  }
  for (int i = 1; i <= m; i += k) {
    cnt = 1ll * a[i] + cnt;
  }
  cout << cnt;
  return 0;
}

```