#include <bits/stdc++.h>
using namespace std;
void FAST_IO() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
}
void err(istream_iterator<string> it) { cerr << endl; }
template <typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
  cerr << *it << ": " << a << " ";
  err(++it, args...);
}
string format(const string& s, ...) {
  va_list args;
  va_start(args, s);
  size_t len = vsnprintf(NULL, 0, s.c_str(), args);
  va_end(args);
  std::vector<char> vec(len + 1);
  va_start(args, s);
  vsnprintf(&vec[0], len + 1, s.c_str(), args);
  va_end(args);
  return &vec[0];
}
const double PI = acos(-1.0);
const int MOD = 1e9 + 7;
const int INF = 2e9;
long long n, k, l;
long long N;
vector<long long> arr;
vector<set<int>> match;
void read() {
  cin >> n >> k >> l;
  N = n * k;
  arr.resize(n * k);
  for (long long i = 0; i < n * k; ++i) {
    cin >> arr[i];
  }
  sort(arr.begin(), arr.end());
}
vector<bool> used;
void solve() {
  int r = upper_bound(arr.begin(), arr.end(), arr[0] + l) - arr.begin();
  --r;
  if (r + 1 < n) {
    cout << 0 << endl;
    return;
  } else {
    long long sum = 0LL;
    if (n == 1)
      sum += arr[0];
    else if (k == 1) {
      for (int i = 0; i < (int)N; ++i) {
        sum += arr[i];
      }
    } else {
      int up = (N - r - 1) / (k - 1);
      int sisa = (N - r - 1) % (k - 1);
      int i;
      for (i = r; i > r - up; --i) {
        sum += arr[i];
      }
      int cnt = 1;
      while (sisa + cnt < k) {
        ++cnt;
        --i;
      }
      sum += arr[i];
      for (int j = 0; j < i; j += k) {
        sum += arr[j];
      }
    }
    cout << sum << endl;
  }
}
int main() {
  FAST_IO();
  int TC = 1;
  for (int tc = (int)1; tc <= (int)TC; ++tc) {
    read();
    solve();
  }
  return 0;
}

### Prompt

Your challenge is to write 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;
void FAST_IO() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
}
void err(istream_iterator<string> it) { cerr << endl; }
template <typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
  cerr << *it << ": " << a << " ";
  err(++it, args...);
}
string format(const string& s, ...) {
  va_list args;
  va_start(args, s);
  size_t len = vsnprintf(NULL, 0, s.c_str(), args);
  va_end(args);
  std::vector<char> vec(len + 1);
  va_start(args, s);
  vsnprintf(&vec[0], len + 1, s.c_str(), args);
  va_end(args);
  return &vec[0];
}
const double PI = acos(-1.0);
const int MOD = 1e9 + 7;
const int INF = 2e9;
long long n, k, l;
long long N;
vector<long long> arr;
vector<set<int>> match;
void read() {
  cin >> n >> k >> l;
  N = n * k;
  arr.resize(n * k);
  for (long long i = 0; i < n * k; ++i) {
    cin >> arr[i];
  }
  sort(arr.begin(), arr.end());
}
vector<bool> used;
void solve() {
  int r = upper_bound(arr.begin(), arr.end(), arr[0] + l) - arr.begin();
  --r;
  if (r + 1 < n) {
    cout << 0 << endl;
    return;
  } else {
    long long sum = 0LL;
    if (n == 1)
      sum += arr[0];
    else if (k == 1) {
      for (int i = 0; i < (int)N; ++i) {
        sum += arr[i];
      }
    } else {
      int up = (N - r - 1) / (k - 1);
      int sisa = (N - r - 1) % (k - 1);
      int i;
      for (i = r; i > r - up; --i) {
        sum += arr[i];
      }
      int cnt = 1;
      while (sisa + cnt < k) {
        ++cnt;
        --i;
      }
      sum += arr[i];
      for (int j = 0; j < i; j += k) {
        sum += arr[j];
      }
    }
    cout << sum << endl;
  }
}
int main() {
  FAST_IO();
  int TC = 1;
  for (int tc = (int)1; tc <= (int)TC; ++tc) {
    read();
    solve();
  }
  return 0;
}

```