#include <bits/stdc++.h>
using namespace std;
map<long long, long long> counter;
std::vector<long long> arr;
long long fastexpo(long long x, long long y, long long m) {
  long long temp = 1;
  while (y > 0) {
    if (y & 1) temp = ((temp % m) * (x % m)) % m;
    x = ((x % m) * (x % m)) % m;
    y >>= 1;
  }
  return temp;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n, k, l, a;
  long long ans = 0, co = 1;
  cin >> n >> k >> l;
  std::vector<long long>::iterator it;
  long long aux = n * k, cur, pos;
  for (int i = 0; i < aux; ++i) {
    cin >> a;
    arr.push_back(a);
    counter[a]++;
  }
  sort(arr.begin(), arr.end());
  if (arr[n - 1] - arr[0] > l) {
    cout << "0";
    return 0;
  }
  it = upper_bound(arr.begin(), arr.end(), arr[0] + l);
  it--;
  pos = it - arr.begin();
  ans += arr[0];
  cur = 0;
  while (co < n) {
    if (pos - (cur + k) >= (n - co - 1)) {
      cur += k;
      ans += arr[cur];
      co++;
      continue;
    }
    while (pos - cur >= n - co) cur++;
    while (co < n) {
      ans += arr[cur];
      cur++;
      co++;
    }
  }
  cout << ans;
  return 0;
}

### Prompt

Your task is to create 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;
map<long long, long long> counter;
std::vector<long long> arr;
long long fastexpo(long long x, long long y, long long m) {
  long long temp = 1;
  while (y > 0) {
    if (y & 1) temp = ((temp % m) * (x % m)) % m;
    x = ((x % m) * (x % m)) % m;
    y >>= 1;
  }
  return temp;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n, k, l, a;
  long long ans = 0, co = 1;
  cin >> n >> k >> l;
  std::vector<long long>::iterator it;
  long long aux = n * k, cur, pos;
  for (int i = 0; i < aux; ++i) {
    cin >> a;
    arr.push_back(a);
    counter[a]++;
  }
  sort(arr.begin(), arr.end());
  if (arr[n - 1] - arr[0] > l) {
    cout << "0";
    return 0;
  }
  it = upper_bound(arr.begin(), arr.end(), arr[0] + l);
  it--;
  pos = it - arr.begin();
  ans += arr[0];
  cur = 0;
  while (co < n) {
    if (pos - (cur + k) >= (n - co - 1)) {
      cur += k;
      ans += arr[cur];
      co++;
      continue;
    }
    while (pos - cur >= n - co) cur++;
    while (co < n) {
      ans += arr[cur];
      cur++;
      co++;
    }
  }
  cout << ans;
  return 0;
}

```