#include <bits/stdc++.h>
using namespace std;
const long long int N = 1e4 + 7;
long long expo(long long x, long long n, long long mod) {
  long long res = 1LL;
  while (n) {
    if (n & 1) res = (res * x) % mod;
    x = (x * x) % mod;
    n = n / 2;
  }
  return res;
}
int main(int argc, char const *argv[]) {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  int i, j;
  int n, k, l, c = 0, x, y;
  long long res = 0, num = INT_MAX, r = 0, temp;
  cin >> n >> k >> l;
  vector<long long> arr(n * k), v;
  for (int i = (0); i < n * k; i++) cin >> arr[i], num = min(num, arr[i]);
  for (int i = (0); i < n * k; i++) {
    c += (arr[i] <= num + l) ? 1 : 0;
  }
  if (c < n) return cout << 0, 0;
  r = 0;
  for (int i = (0); i < n * k; i++) {
    if (arr[i] <= num + l) v.push_back(arr[i]);
  }
  sort(v.begin(), v.end());
  if (k == 1) {
    for (int i = (0); i < (int)((v).size()); i++) r += v[i];
    return cout << r, 0;
  }
  res = (n * k - (int)((v).size())) / (k - 1);
  temp = (n * k - (int)((v).size())) % (k - 1);
  while (res--) {
    r += v.back();
    v.pop_back();
  }
  x = k;
  x -= temp;
  for (int i = (0); i < x; i++) {
    if ((int)((v).size())) {
      if (x - 1 == i) r += v.back();
      v.pop_back();
    }
  }
  i = 0;
  while (i < (int)((v).size())) {
    r += v[i];
    i += k;
  }
  cout << r;
  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;
const long long int N = 1e4 + 7;
long long expo(long long x, long long n, long long mod) {
  long long res = 1LL;
  while (n) {
    if (n & 1) res = (res * x) % mod;
    x = (x * x) % mod;
    n = n / 2;
  }
  return res;
}
int main(int argc, char const *argv[]) {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  int i, j;
  int n, k, l, c = 0, x, y;
  long long res = 0, num = INT_MAX, r = 0, temp;
  cin >> n >> k >> l;
  vector<long long> arr(n * k), v;
  for (int i = (0); i < n * k; i++) cin >> arr[i], num = min(num, arr[i]);
  for (int i = (0); i < n * k; i++) {
    c += (arr[i] <= num + l) ? 1 : 0;
  }
  if (c < n) return cout << 0, 0;
  r = 0;
  for (int i = (0); i < n * k; i++) {
    if (arr[i] <= num + l) v.push_back(arr[i]);
  }
  sort(v.begin(), v.end());
  if (k == 1) {
    for (int i = (0); i < (int)((v).size()); i++) r += v[i];
    return cout << r, 0;
  }
  res = (n * k - (int)((v).size())) / (k - 1);
  temp = (n * k - (int)((v).size())) % (k - 1);
  while (res--) {
    r += v.back();
    v.pop_back();
  }
  x = k;
  x -= temp;
  for (int i = (0); i < x; i++) {
    if ((int)((v).size())) {
      if (x - 1 == i) r += v.back();
      v.pop_back();
    }
  }
  i = 0;
  while (i < (int)((v).size())) {
    r += v[i];
    i += k;
  }
  cout << r;
  return 0;
}

```