#include <bits/stdc++.h>
using namespace std;
long long gcd(long long a, long long b) { return (b ? gcd(b, a % b) : a); }
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
long long inf = 1e9 + 7;
long long abso(long long x) {
  if (x > 0) return x;
  return -1 * x;
}
int main() {
  std::ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n, k, l;
  cin >> n >> k >> l;
  long long m = n * k;
  vector<long long> arr(m);
  for (long long i = 0; i < m; i++) cin >> arr[i];
  sort(arr.begin(), arr.end());
  if (arr[n - 1] > arr[0] + l) {
    cout << 0;
    return 0;
  }
  long long ans = 0;
  long long ind =
      upper_bound(arr.begin(), arr.end(), arr[0] + l) - (arr.begin() + 1);
  long long cnt = 0;
  while (cnt < n && (m - 1) - ind >= k) {
    ans += arr[ind];
    m -= k;
    ind--;
    cnt++;
  }
  long long i = 0;
  while (cnt < n) {
    ans += arr[i];
    cnt++;
    i += k;
  }
  cout << ans;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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 gcd(long long a, long long b) { return (b ? gcd(b, a % b) : a); }
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
long long inf = 1e9 + 7;
long long abso(long long x) {
  if (x > 0) return x;
  return -1 * x;
}
int main() {
  std::ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n, k, l;
  cin >> n >> k >> l;
  long long m = n * k;
  vector<long long> arr(m);
  for (long long i = 0; i < m; i++) cin >> arr[i];
  sort(arr.begin(), arr.end());
  if (arr[n - 1] > arr[0] + l) {
    cout << 0;
    return 0;
  }
  long long ans = 0;
  long long ind =
      upper_bound(arr.begin(), arr.end(), arr[0] + l) - (arr.begin() + 1);
  long long cnt = 0;
  while (cnt < n && (m - 1) - ind >= k) {
    ans += arr[ind];
    m -= k;
    ind--;
    cnt++;
  }
  long long i = 0;
  while (cnt < n) {
    ans += arr[i];
    cnt++;
    i += k;
  }
  cout << ans;
}

```