#include <bits/stdc++.h>
using namespace std;
const int N = 1E5 + 10;
int num[N];
int n, k, m, xx;
int main() {
  scanf("%d%d%d", &n, &k, &xx);
  m = n * k;
  for (int i = 0; i < m; i++) scanf("%d", &num[i]);
  sort(num, num + m);
  int up = num[0] + xx;
  int l = 0, r = m - 1;
  while (l < r) {
    int mid = (l + r) >> 1;
    if (num[mid] > up)
      r = mid - 1;
    else
      l = mid + 1;
  }
  while (num[l] > up) l--;
  if (l + 1 < n) {
    puts("0");
    return 0;
  }
  int left = l + 1 - n;
  int cnt;
  if (k == 1)
    cnt = 0;
  else
    cnt = left / (k - 1) + 1;
  long long tot = 0;
  int op = 0;
  for (int i = 0; i < cnt; i++) {
    tot += (long long)num[op];
    op += k;
  }
  cnt = n - cnt;
  op = l;
  for (int i = 0; i < cnt; i++) {
    tot += (long long)num[op];
    op--;
  }
  printf("%lld", tot);
}

### Prompt

Please create a solution in cpp 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 N = 1E5 + 10;
int num[N];
int n, k, m, xx;
int main() {
  scanf("%d%d%d", &n, &k, &xx);
  m = n * k;
  for (int i = 0; i < m; i++) scanf("%d", &num[i]);
  sort(num, num + m);
  int up = num[0] + xx;
  int l = 0, r = m - 1;
  while (l < r) {
    int mid = (l + r) >> 1;
    if (num[mid] > up)
      r = mid - 1;
    else
      l = mid + 1;
  }
  while (num[l] > up) l--;
  if (l + 1 < n) {
    puts("0");
    return 0;
  }
  int left = l + 1 - n;
  int cnt;
  if (k == 1)
    cnt = 0;
  else
    cnt = left / (k - 1) + 1;
  long long tot = 0;
  int op = 0;
  for (int i = 0; i < cnt; i++) {
    tot += (long long)num[op];
    op += k;
  }
  cnt = n - cnt;
  op = l;
  for (int i = 0; i < cnt; i++) {
    tot += (long long)num[op];
    op--;
  }
  printf("%lld", tot);
}

```