#include <bits/stdc++.h>
using namespace std;
const int maxN = 100000, intMax = 2147483647, maxK = 100000, maxM = 100000;
int min(int a, int b) {
  if (a < b) return a;
  return b;
}
long long n, k, l, a[maxM];
int main() {
  scanf("%lld %lld %lld", &n, &k, &l);
  for (long long i = 0; i < n * k; i++) scanf("%lld", &a[i]);
  sort(begin(a), begin(a) + n * k);
  if (a[n - 1] - a[0] > l) {
    printf("0");
    return 0;
  }
  long long i = 0;
  for (; i < n * k && a[i] - a[0] <= l; i++)
    ;
  long long s = 0;
  long long ind = 0;
  long long q = 0;
  while (i - ind - k >= n - q - 1 && q < n - 1) {
    s += a[ind];
    ind += k;
    q++;
  }
  s += a[ind];
  q++;
  ind = i - 1;
  for (; q < n; q++, ind--) s += a[ind];
  printf("%lld", s);
  return 0;
}

### Prompt

Generate 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 maxN = 100000, intMax = 2147483647, maxK = 100000, maxM = 100000;
int min(int a, int b) {
  if (a < b) return a;
  return b;
}
long long n, k, l, a[maxM];
int main() {
  scanf("%lld %lld %lld", &n, &k, &l);
  for (long long i = 0; i < n * k; i++) scanf("%lld", &a[i]);
  sort(begin(a), begin(a) + n * k);
  if (a[n - 1] - a[0] > l) {
    printf("0");
    return 0;
  }
  long long i = 0;
  for (; i < n * k && a[i] - a[0] <= l; i++)
    ;
  long long s = 0;
  long long ind = 0;
  long long q = 0;
  while (i - ind - k >= n - q - 1 && q < n - 1) {
    s += a[ind];
    ind += k;
    q++;
  }
  s += a[ind];
  q++;
  ind = i - 1;
  for (; q < n; q++, ind--) s += a[ind];
  printf("%lld", s);
  return 0;
}

```