#include <bits/stdc++.h>
using namespace std;
template <class c>
struct rge {
  c b, e;
};
template <class c>
rge<c> range(c i, c j) {
  return rge<c>{i, j};
}
template <class c>
auto dud(c* x) -> decltype(cerr << *x, 0);
template <class c>
char dud(...);
struct debug {
  template <class c>
  debug& operator<<(const c&) {
    return *this;
  }
};
const long long int N = 2e5 + 5;
long long int mod = 1e9 + 7;
long long int dx[4] = {0, 0, +1, -1};
long long int dy[4] = {+1, -1, 0, 0};
void solve() {
  long long int n, k, l;
  cin >> n >> k >> l;
  long long int n1 = n * k;
  vector<long long int> v(n1, 0);
  for (long long int i = 0; i < n1; i++) cin >> v[i];
  sort((v).begin(), (v).end());
  long long int ind = upper_bound((v).begin(), (v).end(), v[0] + l) - v.begin();
  if (ind < n) {
    cout << 0 << "\n";
    return;
  }
  long long int ans = 0, u = 0;
  for (long long int i = 0; i < n; i++) {
    ans += v[u++];
    for (long long int j = 0; j < k - 1; j++) {
      if (ind - u > n - i - 1)
        u++;
      else
        break;
    }
  }
  cout << ans << "\n";
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long int t = 1;
  while (t--) solve();
  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;
template <class c>
struct rge {
  c b, e;
};
template <class c>
rge<c> range(c i, c j) {
  return rge<c>{i, j};
}
template <class c>
auto dud(c* x) -> decltype(cerr << *x, 0);
template <class c>
char dud(...);
struct debug {
  template <class c>
  debug& operator<<(const c&) {
    return *this;
  }
};
const long long int N = 2e5 + 5;
long long int mod = 1e9 + 7;
long long int dx[4] = {0, 0, +1, -1};
long long int dy[4] = {+1, -1, 0, 0};
void solve() {
  long long int n, k, l;
  cin >> n >> k >> l;
  long long int n1 = n * k;
  vector<long long int> v(n1, 0);
  for (long long int i = 0; i < n1; i++) cin >> v[i];
  sort((v).begin(), (v).end());
  long long int ind = upper_bound((v).begin(), (v).end(), v[0] + l) - v.begin();
  if (ind < n) {
    cout << 0 << "\n";
    return;
  }
  long long int ans = 0, u = 0;
  for (long long int i = 0; i < n; i++) {
    ans += v[u++];
    for (long long int j = 0; j < k - 1; j++) {
      if (ind - u > n - i - 1)
        u++;
      else
        break;
    }
  }
  cout << ans << "\n";
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long int t = 1;
  while (t--) solve();
  return 0;
}

```