#include <bits/stdc++.h>
using namespace std;
long long int power(long long int a, long long int b) {
  long long int res = 1;
  a %= 1000000007;
  assert(b >= 0);
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % 1000000007;
    a = a * a % 1000000007;
  }
  return res;
}
long long int modInverse(long long int a) { return power(a, 1000000007 - 2); }
const int N = 500023;
bool vis[N];
vector<int> adj[N];
void solve() {
  long long int n, k, l;
  cin >> n >> k >> l;
  vector<long long int> v;
  long long int m = n * k;
  multiset<long long int> s;
  for (long long int i = 0; i < m; i++) {
    long long int c;
    cin >> c;
    v.push_back(c);
    s.insert(c);
  }
  sort((v).begin(), (v).end());
  if ((v[n - 1] - v[0]) > l) {
    cout << 0;
    return;
  }
  long long int ans = 0;
  while (true) {
    if (s.empty()) break;
    for (int i = 1; i < k; i++) {
      auto it = s.end();
      it--;
      s.erase(it);
    }
    auto it = s.lower_bound(v[0] + l + 1);
    it--;
    ans += (*it);
    s.erase(it);
  }
  cout << ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int T = 1;
  int t = 0;
  while (t++ < T) {
    solve();
  }
  cerr << "Time : " << 1000 * ((double)clock()) / (double)CLOCKS_PER_SEC
       << "ms\n";
}

### 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;
long long int power(long long int a, long long int b) {
  long long int res = 1;
  a %= 1000000007;
  assert(b >= 0);
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % 1000000007;
    a = a * a % 1000000007;
  }
  return res;
}
long long int modInverse(long long int a) { return power(a, 1000000007 - 2); }
const int N = 500023;
bool vis[N];
vector<int> adj[N];
void solve() {
  long long int n, k, l;
  cin >> n >> k >> l;
  vector<long long int> v;
  long long int m = n * k;
  multiset<long long int> s;
  for (long long int i = 0; i < m; i++) {
    long long int c;
    cin >> c;
    v.push_back(c);
    s.insert(c);
  }
  sort((v).begin(), (v).end());
  if ((v[n - 1] - v[0]) > l) {
    cout << 0;
    return;
  }
  long long int ans = 0;
  while (true) {
    if (s.empty()) break;
    for (int i = 1; i < k; i++) {
      auto it = s.end();
      it--;
      s.erase(it);
    }
    auto it = s.lower_bound(v[0] + l + 1);
    it--;
    ans += (*it);
    s.erase(it);
  }
  cout << ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int T = 1;
  int t = 0;
  while (t++ < T) {
    solve();
  }
  cerr << "Time : " << 1000 * ((double)clock()) / (double)CLOCKS_PER_SEC
       << "ms\n";
}

```