#include <bits/stdc++.h>
using namespace std;
long long int a[100005];
long long int b[100005];
int cmp(long long int a, long long int b) { return a < b; }
int main() {
  long long int i, j, m, n, k, l, min, sum, num;
  scanf("%lld%lld%lld", &n, &k, &l);
  m = n * k;
  sum = 0;
  min = 10000000000;
  for (i = 1; i <= m; i++) {
    scanf("%lld", &a[i]);
    if (min > a[i]) min = a[i];
  }
  j = 0;
  for (i = 1; i <= m; i++) {
    if (a[i] <= min + l) {
      j++;
      b[j] = a[i];
    }
  }
  num = j;
  if (num < n) {
    printf("0");
    return 0;
  }
  sort(b + 1, b + num + 1, cmp);
  i = 1;
  j = 0;
  while (i <= num) {
    if (num - i + 1 - (n - j) >= k) {
      sum = sum + b[i];
      j++;
      i = i + k;
    } else if (num - i + 1 > n - j) {
      sum = sum + b[i];
      j++;
      i = i + num - i + 1 - (n - j);
    } else {
      sum = sum + b[i];
      j++;
      i++;
    }
  }
  printf("%lld", sum);
  return 0;
}

### 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>
using namespace std;
long long int a[100005];
long long int b[100005];
int cmp(long long int a, long long int b) { return a < b; }
int main() {
  long long int i, j, m, n, k, l, min, sum, num;
  scanf("%lld%lld%lld", &n, &k, &l);
  m = n * k;
  sum = 0;
  min = 10000000000;
  for (i = 1; i <= m; i++) {
    scanf("%lld", &a[i]);
    if (min > a[i]) min = a[i];
  }
  j = 0;
  for (i = 1; i <= m; i++) {
    if (a[i] <= min + l) {
      j++;
      b[j] = a[i];
    }
  }
  num = j;
  if (num < n) {
    printf("0");
    return 0;
  }
  sort(b + 1, b + num + 1, cmp);
  i = 1;
  j = 0;
  while (i <= num) {
    if (num - i + 1 - (n - j) >= k) {
      sum = sum + b[i];
      j++;
      i = i + k;
    } else if (num - i + 1 > n - j) {
      sum = sum + b[i];
      j++;
      i = i + num - i + 1 - (n - j);
    } else {
      sum = sum + b[i];
      j++;
      i++;
    }
  }
  printf("%lld", sum);
  return 0;
}

```