#include <bits/stdc++.h>
void kuaipai(long long *left, long long *right) {
  long long key = *left;
  long long *stl = left;
  long long *str = right;
  if (left >= right) return;
  while (left < right) {
    while (left < right && key <= *right) {
      right--;
    }
    *left = *right;
    while (left < right && key >= *left) {
      left++;
    }
    *right = *left;
  }
  *left = key;
  kuaipai(stl, left - 1);
  kuaipai(right + 1, str);
}
int main() {
  int n, k, l;
  long long sum = 0;
  int i, j, t = 0;
  long long a[100001] = {0};
  scanf("%d %d %d", &n, &k, &l);
  for (i = 0; i < n * k; i++) scanf("%I64d", &a[i]);
  kuaipai(&a[0], &a[i - 1]);
  a[i + 1] = a[i] + 1;
  if (a[n - 1] - a[0] > l)
    printf("0\n");
  else {
    for (i = 0; (a[i] > a[0] + l || a[i + 1] <= a[0] + l) && i < n * k; i++)
      ;
    for (j = 0; j <= i; j = j + k) {
      sum = sum + a[j];
      a[j] = 0;
      t++;
    }
    j = i + 1;
    while (t < n) {
      j--;
      if (a[j] == 0) continue;
      sum = sum + a[j];
      t++;
    }
    printf("%I64d\n", sum);
  }
  return 0;
}

### Prompt

Your task is to create 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>
void kuaipai(long long *left, long long *right) {
  long long key = *left;
  long long *stl = left;
  long long *str = right;
  if (left >= right) return;
  while (left < right) {
    while (left < right && key <= *right) {
      right--;
    }
    *left = *right;
    while (left < right && key >= *left) {
      left++;
    }
    *right = *left;
  }
  *left = key;
  kuaipai(stl, left - 1);
  kuaipai(right + 1, str);
}
int main() {
  int n, k, l;
  long long sum = 0;
  int i, j, t = 0;
  long long a[100001] = {0};
  scanf("%d %d %d", &n, &k, &l);
  for (i = 0; i < n * k; i++) scanf("%I64d", &a[i]);
  kuaipai(&a[0], &a[i - 1]);
  a[i + 1] = a[i] + 1;
  if (a[n - 1] - a[0] > l)
    printf("0\n");
  else {
    for (i = 0; (a[i] > a[0] + l || a[i + 1] <= a[0] + l) && i < n * k; i++)
      ;
    for (j = 0; j <= i; j = j + k) {
      sum = sum + a[j];
      a[j] = 0;
      t++;
    }
    j = i + 1;
    while (t < n) {
      j--;
      if (a[j] == 0) continue;
      sum = sum + a[j];
      t++;
    }
    printf("%I64d\n", sum);
  }
  return 0;
}

```