#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
template <class T>
bool chmax(T& a, const T& b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
bool chmin(T& a, const T& b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class S, class T>
ostream& operator<<(ostream& os, const pair<S, T> v) {
  os << "(" << v.first << ", " << v.second << ")";
  return os;
}
template <class T>
ostream& operator<<(ostream& os, const vector<T> v) {
  for (int i = 0; i < (int)v.size(); i++) {
    if (i > 0) {
      os << " ";
    }
    os << v[i];
  }
  return os;
}
template <class T>
ostream& operator<<(ostream& os, const vector<vector<T>> v) {
  for (int i = 0; i < (int)v.size(); i++) {
    if (i > 0) {
      os << endl;
    }
    os << v[i];
  }
  return os;
}
string num2bit(long long num, long long len) {
  string bit = "";
  for (int i = 0; i < (int)len; ++i) {
    bit += char('0' + (num >> i & 1));
  }
  return bit;
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  long long n, k, l;
  cin >> n >> k >> l;
  long long m = n * k;
  vector<long long> a(m);
  for (int i = 0; i < (int)m; ++i) cin >> a[i];
  sort(a.rbegin(), a.rend());
  long long res = 0;
  long long tot = 0;
  for (int i = 0; i < (int)m; ++i) {
    if (a[i] > a[m - 1] + l) continue;
    if (i + 1 >= (tot + 1) * k) {
      res += a[i];
      tot++;
    }
  }
  if (tot == n) {
    cout << res << endl;
  } else {
    cout << 0 << endl;
  }
  return 0;
}

### Prompt

In CPP, your task is to solve 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 long long mod = 1e9 + 7;
template <class T>
bool chmax(T& a, const T& b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
bool chmin(T& a, const T& b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class S, class T>
ostream& operator<<(ostream& os, const pair<S, T> v) {
  os << "(" << v.first << ", " << v.second << ")";
  return os;
}
template <class T>
ostream& operator<<(ostream& os, const vector<T> v) {
  for (int i = 0; i < (int)v.size(); i++) {
    if (i > 0) {
      os << " ";
    }
    os << v[i];
  }
  return os;
}
template <class T>
ostream& operator<<(ostream& os, const vector<vector<T>> v) {
  for (int i = 0; i < (int)v.size(); i++) {
    if (i > 0) {
      os << endl;
    }
    os << v[i];
  }
  return os;
}
string num2bit(long long num, long long len) {
  string bit = "";
  for (int i = 0; i < (int)len; ++i) {
    bit += char('0' + (num >> i & 1));
  }
  return bit;
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  long long n, k, l;
  cin >> n >> k >> l;
  long long m = n * k;
  vector<long long> a(m);
  for (int i = 0; i < (int)m; ++i) cin >> a[i];
  sort(a.rbegin(), a.rend());
  long long res = 0;
  long long tot = 0;
  for (int i = 0; i < (int)m; ++i) {
    if (a[i] > a[m - 1] + l) continue;
    if (i + 1 >= (tot + 1) * k) {
      res += a[i];
      tot++;
    }
  }
  if (tot == n) {
    cout << res << endl;
  } else {
    cout << 0 << endl;
  }
  return 0;
}

```