#include <bits/stdc++.h>
using namespace std;
template <class T>
inline bool rd(T &num) {
  char in;
  bool IsN = false;
  while (in = getchar(), in != '-' && (in < '0' || in > '9'))
    if (in == EOF) return false;
  in == '-' ? (IsN = true, num = 0) : num = in - '0';
  while (in = getchar(), in >= '0' && in <= '9') num = num * 10 + in - '0';
  return IsN ? num = -num : num, true;
}
string nico;
const int maxn = 2e5 + 7;
const int inf = 2e9 + 7;
const int mod = 1e9 + 7;
int len[maxn];
int main() {
  ios::sync_with_stdio(false);
  long long n, k, x;
  cin >> n >> k >> x;
  for (register int i = (0); i < (n * k); i++) cin >> len[i];
  sort(len, len + n * k);
  int l = 0, r = 0;
  for (register int i = (0); i < (n * k); i++) {
    if (len[i] - len[0] <= x) r = i;
  };
  ;
  ;
  ;
  if (r + 1 < n) {
    cout << 0 << endl;
    return 0;
  }
  long long ans = 0;
  int cnt = 0;
  int have = n * k - r - 1;
  while (have) {
    cnt++;
    ans += len[r];
    have -= k - 1;
    if (have < 0) {
      ans -= len[r];
      r -= abs(have);
      have = 0;
      ans += len[r];
    }
    r--;
  }
  while (cnt != n) {
    cnt++;
    ans += len[r - k + 1];
    r -= k;
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Create a solution in Cpp for 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;
template <class T>
inline bool rd(T &num) {
  char in;
  bool IsN = false;
  while (in = getchar(), in != '-' && (in < '0' || in > '9'))
    if (in == EOF) return false;
  in == '-' ? (IsN = true, num = 0) : num = in - '0';
  while (in = getchar(), in >= '0' && in <= '9') num = num * 10 + in - '0';
  return IsN ? num = -num : num, true;
}
string nico;
const int maxn = 2e5 + 7;
const int inf = 2e9 + 7;
const int mod = 1e9 + 7;
int len[maxn];
int main() {
  ios::sync_with_stdio(false);
  long long n, k, x;
  cin >> n >> k >> x;
  for (register int i = (0); i < (n * k); i++) cin >> len[i];
  sort(len, len + n * k);
  int l = 0, r = 0;
  for (register int i = (0); i < (n * k); i++) {
    if (len[i] - len[0] <= x) r = i;
  };
  ;
  ;
  ;
  if (r + 1 < n) {
    cout << 0 << endl;
    return 0;
  }
  long long ans = 0;
  int cnt = 0;
  int have = n * k - r - 1;
  while (have) {
    cnt++;
    ans += len[r];
    have -= k - 1;
    if (have < 0) {
      ans -= len[r];
      r -= abs(have);
      have = 0;
      ans += len[r];
    }
    r--;
  }
  while (cnt != n) {
    cnt++;
    ans += len[r - k + 1];
    r -= k;
  }
  cout << ans << endl;
  return 0;
}

```