#include <bits/stdc++.h>
using namespace std;
map<long long int, pair<int, int> > M;
map<long long int, pair<int, int> >::iterator it1, it2;
long long int arr[100005];
pair<long long int, long long int> a[100005];
long long int dum[100005];
int main() {
  long long int n, k;
  long long int l;
  cin >> n >> k >> l;
  long long int i, j, kk = 1;
  for (i = 1; i <= n * k; i++) {
    cin >> dum[i];
  }
  sort(dum + 1, dum + (n * k) + 1);
  for (i = 1; i <= n * k; i++) {
    if (M[dum[i]].first == 0) {
      M[dum[i]].first = 1;
      M[dum[i]].second = kk++;
    } else
      M[dum[i]].first++;
  }
  for (it1 = M.begin(); it1 != M.end(); it1++) {
    a[it1->second.second].first = it1->first;
    a[it1->second.second].second =
        a[it1->second.second - 1].second + it1->second.first;
  }
  it2 = M.begin();
  it1 = M.upper_bound(it2->first + l);
  it1--;
  int ind = it1->second.second;
  if (l == 0) {
    it1 = M.end();
    it1--;
    if (it1 != M.begin()) {
      cout << 0 << "\n";
      return 0;
    } else {
      cout << (a[1].second / k) * a[1].first << "\n";
      return 0;
    }
    exit(0);
  }
  if (k == 1) {
    it1 = M.end();
    it1--;
    it2 = M.begin();
    if (it2->first + l < it1->first)
      cout << 0 << "\n";
    else {
      long long int ans = 0;
      for (it1 = M.begin(); it1 != M.end(); it1++) {
        ans += it1->second.first * it1->first;
      }
      cout << ans;
      exit(0);
    }
    exit(0);
  }
  if ((a[ind].second) * (k - 1) < (n * k - a[ind].second)) {
    cout << 0;
    exit(0);
  }
  long long int ans = 0;
  long long int rem = n * k - a[ind].second;
  j = ind;
  while (M[a[j].first].first * (k - 1) < rem) {
    ans += M[a[j].first].first * a[j].first;
    rem -= M[a[j].first].first * (k - 1);
    j--;
  }
  while (j > 0) {
    if (rem + M[a[j].first].first >= k) {
      ans += a[j].first;
      M[a[j].first].first -= (k - rem);
      rem = 0;
    } else {
      rem += M[a[j].first].first;
      j--;
    }
  }
  cout << ans;
}

### Prompt

Please provide a Cpp coded solution 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;
map<long long int, pair<int, int> > M;
map<long long int, pair<int, int> >::iterator it1, it2;
long long int arr[100005];
pair<long long int, long long int> a[100005];
long long int dum[100005];
int main() {
  long long int n, k;
  long long int l;
  cin >> n >> k >> l;
  long long int i, j, kk = 1;
  for (i = 1; i <= n * k; i++) {
    cin >> dum[i];
  }
  sort(dum + 1, dum + (n * k) + 1);
  for (i = 1; i <= n * k; i++) {
    if (M[dum[i]].first == 0) {
      M[dum[i]].first = 1;
      M[dum[i]].second = kk++;
    } else
      M[dum[i]].first++;
  }
  for (it1 = M.begin(); it1 != M.end(); it1++) {
    a[it1->second.second].first = it1->first;
    a[it1->second.second].second =
        a[it1->second.second - 1].second + it1->second.first;
  }
  it2 = M.begin();
  it1 = M.upper_bound(it2->first + l);
  it1--;
  int ind = it1->second.second;
  if (l == 0) {
    it1 = M.end();
    it1--;
    if (it1 != M.begin()) {
      cout << 0 << "\n";
      return 0;
    } else {
      cout << (a[1].second / k) * a[1].first << "\n";
      return 0;
    }
    exit(0);
  }
  if (k == 1) {
    it1 = M.end();
    it1--;
    it2 = M.begin();
    if (it2->first + l < it1->first)
      cout << 0 << "\n";
    else {
      long long int ans = 0;
      for (it1 = M.begin(); it1 != M.end(); it1++) {
        ans += it1->second.first * it1->first;
      }
      cout << ans;
      exit(0);
    }
    exit(0);
  }
  if ((a[ind].second) * (k - 1) < (n * k - a[ind].second)) {
    cout << 0;
    exit(0);
  }
  long long int ans = 0;
  long long int rem = n * k - a[ind].second;
  j = ind;
  while (M[a[j].first].first * (k - 1) < rem) {
    ans += M[a[j].first].first * a[j].first;
    rem -= M[a[j].first].first * (k - 1);
    j--;
  }
  while (j > 0) {
    if (rem + M[a[j].first].first >= k) {
      ans += a[j].first;
      M[a[j].first].first -= (k - rem);
      rem = 0;
    } else {
      rem += M[a[j].first].first;
      j--;
    }
  }
  cout << ans;
}

```