#include <bits/stdc++.h>
using namespace std;
const int HashMod = 1125022017;
const int dx[4] = {0, 1, 0, -1};
const int dy[4] = {1, 0, -1, 0};
const int oo = 2e9;
const int nmax = 1e5 + 9;
int n, k, l;
int a[nmax];
vector<int> b;
void Inp() {
  ios::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  cin >> n >> k >> l;
  for (int i = 1; i <= n * k; ++i) cin >> a[i];
  sort(a + 1, a + n * k + 1);
}
bool comp(int x, int y) { return x > y; }
void Process() {
  int firs = 1;
  int last = upper_bound(a + 1, a + n * k + 1, a[1] + l) - a - 1;
  if (last - firs + 1 < n)
    cout << 0;
  else {
    long long ans = 0;
    int d = n;
    for (int i = 1; i <= last; ++i)
      if (i % k == 1) {
        ans = ans + a[i], --d;
        if (d == 0) break;
      } else
        b.push_back(a[i]);
    if (d > 0) {
      sort(b.begin(), b.end(), comp);
      for (int i = 0; i < d; i++) ans += b[i];
    }
    cout << ans;
  }
}
int main() {
  Inp();
  Process();
  return 0;
}

### 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;
const int HashMod = 1125022017;
const int dx[4] = {0, 1, 0, -1};
const int dy[4] = {1, 0, -1, 0};
const int oo = 2e9;
const int nmax = 1e5 + 9;
int n, k, l;
int a[nmax];
vector<int> b;
void Inp() {
  ios::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  cin >> n >> k >> l;
  for (int i = 1; i <= n * k; ++i) cin >> a[i];
  sort(a + 1, a + n * k + 1);
}
bool comp(int x, int y) { return x > y; }
void Process() {
  int firs = 1;
  int last = upper_bound(a + 1, a + n * k + 1, a[1] + l) - a - 1;
  if (last - firs + 1 < n)
    cout << 0;
  else {
    long long ans = 0;
    int d = n;
    for (int i = 1; i <= last; ++i)
      if (i % k == 1) {
        ans = ans + a[i], --d;
        if (d == 0) break;
      } else
        b.push_back(a[i]);
    if (d > 0) {
      sort(b.begin(), b.end(), comp);
      for (int i = 0; i < d; i++) ans += b[i];
    }
    cout << ans;
  }
}
int main() {
  Inp();
  Process();
  return 0;
}

```