#include <bits/stdc++.h>
using namespace std;
long long n, k, l;
long long len[100005];
int vis[100005];
int main() {
  while (~scanf("%lld %lld %lld", &n, &k, &l)) {
    long long m = n * k;
    for (long long i = 0; i < m; i++) scanf("%lld", &len[i]);
    sort(len, len + m);
    int t = upper_bound(len, len + m, len[0] + l) - len;
    if (n / k == 1 || t < n) {
      cout << 0 << endl;
      continue;
    }
    memset(vis, 0, sizeof vis);
    int idx = 0;
    stack<int> que;
    for (int i = 1; i <= n; i++) que.push(t - i);
    long long ans = 0, flag = 0;
    for (int i = 0; i < n; i++) {
      if (flag)
        ans += len[que.top()];
      else
        ans += len[idx];
      que.pop();
      for (int j = 0; j < k; j++) {
        if (que.size() && idx != que.top())
          idx++;
        else if (flag == 0)
          flag = 1, idx = t;
        else
          idx++;
      }
    }
    cout << ans << endl;
  }
}

### Prompt

Please create a solution in cpp 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 n, k, l;
long long len[100005];
int vis[100005];
int main() {
  while (~scanf("%lld %lld %lld", &n, &k, &l)) {
    long long m = n * k;
    for (long long i = 0; i < m; i++) scanf("%lld", &len[i]);
    sort(len, len + m);
    int t = upper_bound(len, len + m, len[0] + l) - len;
    if (n / k == 1 || t < n) {
      cout << 0 << endl;
      continue;
    }
    memset(vis, 0, sizeof vis);
    int idx = 0;
    stack<int> que;
    for (int i = 1; i <= n; i++) que.push(t - i);
    long long ans = 0, flag = 0;
    for (int i = 0; i < n; i++) {
      if (flag)
        ans += len[que.top()];
      else
        ans += len[idx];
      que.pop();
      for (int j = 0; j < k; j++) {
        if (que.size() && idx != que.top())
          idx++;
        else if (flag == 0)
          flag = 1, idx = t;
        else
          idx++;
      }
    }
    cout << ans << endl;
  }
}

```