#include <bits/stdc++.h>
using namespace std;
const long long INF = 0x3f3f3f3f;
const long long MOD = 1e9 + 7ll;
const long long MAX = 9e6 + 10;
long long n, k, l;
long long a[MAX];
int main() {
  while (~scanf("%I64d%I64d%I64d", &n, &k, &l)) {
    for (long long i = 1; i <= n * k; i++) scanf("%I64d", &a[i]);
    long long tol = n * k;
    sort(a + 1, a + 1 + tol);
    long long mi = a[1];
    long long ma = a[1] + l;
    if (a[n] - a[1] > l) {
      printf("0\n");
      continue;
    }
    long long pos = -1;
    for (long long i = tol; i > 0; i--)
      if (a[i] <= ma) {
        pos = i;
        break;
      }
    long long ans = 0;
    long long i = tol;
    long long j = pos;
    long long num = 0;
    while (i - (k - 1) > pos) {
      ans += a[j];
      j--;
      i -= (k - 1);
      num++;
    }
    for (long long i = 1; i <= pos - num; i += k) ans += a[i];
    printf("%I64d\n", ans);
  }
  return 0;
}

### 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 long long INF = 0x3f3f3f3f;
const long long MOD = 1e9 + 7ll;
const long long MAX = 9e6 + 10;
long long n, k, l;
long long a[MAX];
int main() {
  while (~scanf("%I64d%I64d%I64d", &n, &k, &l)) {
    for (long long i = 1; i <= n * k; i++) scanf("%I64d", &a[i]);
    long long tol = n * k;
    sort(a + 1, a + 1 + tol);
    long long mi = a[1];
    long long ma = a[1] + l;
    if (a[n] - a[1] > l) {
      printf("0\n");
      continue;
    }
    long long pos = -1;
    for (long long i = tol; i > 0; i--)
      if (a[i] <= ma) {
        pos = i;
        break;
      }
    long long ans = 0;
    long long i = tol;
    long long j = pos;
    long long num = 0;
    while (i - (k - 1) > pos) {
      ans += a[j];
      j--;
      i -= (k - 1);
      num++;
    }
    for (long long i = 1; i <= pos - num; i += k) ans += a[i];
    printf("%I64d\n", ans);
  }
  return 0;
}

```