#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 5;
int a[maxn];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int i, j, k, m, n, tn;
  cin >> n >> k >> m;
  tn = n;
  for (int i = 1; i <= n * k; ++i) cin >> a[i];
  n = n * k;
  sort(a + 1, a + n + 1);
  int mx = a[1] + m;
  int pos = upper_bound(a + 1, a + n + 1, mx) - a - 1;
  if (pos < tn) {
    cout << "0" << endl;
    return 0;
  }
  vector<int> av;
  int cnt = 0;
  for (int i = 1; i <= n; ++i) {
    if (a[i] <= mx)
      av.push_back(a[i]);
    else
      cnt++;
  }
  reverse(av.begin(), av.end());
  long long ans = 0;
  queue<int> q;
  for (auto it : av) {
    q.push(it);
  }
  while (!q.empty()) {
    auto it = q.front();
    q.pop();
    int rem = k - 1;
    int use = min(rem, cnt);
    cnt -= use;
    rem -= use;
    int tans = it;
    while (rem) {
      auto pt = q.front();
      q.pop();
      tans = min(tans, pt);
      rem--;
    }
    ans += tans;
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Please formulate 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 = 1e5 + 5;
int a[maxn];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int i, j, k, m, n, tn;
  cin >> n >> k >> m;
  tn = n;
  for (int i = 1; i <= n * k; ++i) cin >> a[i];
  n = n * k;
  sort(a + 1, a + n + 1);
  int mx = a[1] + m;
  int pos = upper_bound(a + 1, a + n + 1, mx) - a - 1;
  if (pos < tn) {
    cout << "0" << endl;
    return 0;
  }
  vector<int> av;
  int cnt = 0;
  for (int i = 1; i <= n; ++i) {
    if (a[i] <= mx)
      av.push_back(a[i]);
    else
      cnt++;
  }
  reverse(av.begin(), av.end());
  long long ans = 0;
  queue<int> q;
  for (auto it : av) {
    q.push(it);
  }
  while (!q.empty()) {
    auto it = q.front();
    q.pop();
    int rem = k - 1;
    int use = min(rem, cnt);
    cnt -= use;
    rem -= use;
    int tans = it;
    while (rem) {
      auto pt = q.front();
      q.pop();
      tans = min(tans, pt);
      rem--;
    }
    ans += tans;
  }
  cout << ans << endl;
  return 0;
}

```