#include <bits/stdc++.h>
using namespace std;
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long power(long long x, long long p, long long MOD) {
  if (p == 0) return 1 % MOD;
  if (p == 1) return x % MOD;
  long long res = power(x, p / 2, MOD);
  res = (long long)res * res % MOD;
  if (p & 1) res = (long long)res * x % MOD;
  return res;
}
int32_t main() {
  long long n, K, L;
  cin >> n >> K >> L;
  vector<long long> p;
  for (long long i = 0; i < n * K; i++) {
    long long x;
    cin >> x;
    p.push_back(x);
  }
  sort((p).begin(), (p).end());
  long long hi = upper_bound((p).begin(), (p).end(), p[0] + L) - p.begin();
  if (hi < n) {
    cout << 0 << endl;
    return 0;
  }
  vector<bool> used(hi);
  long long tot = 0;
  for (long long i = 0; i < hi; i += K) {
    used[i] = true;
    tot++;
  }
  long long ptr = hi - 1;
  while (tot < n) {
    assert(ptr >= 0);
    if (!used[ptr]) used[ptr] = true, tot++;
    ptr--;
  }
  long long ans = 0;
  for (long long i = 0; i < hi; i++) {
    if (used[i]) ans += p[i];
  }
  cout << ans << endl;
}

### Prompt

Develop a solution in cpp 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;
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long power(long long x, long long p, long long MOD) {
  if (p == 0) return 1 % MOD;
  if (p == 1) return x % MOD;
  long long res = power(x, p / 2, MOD);
  res = (long long)res * res % MOD;
  if (p & 1) res = (long long)res * x % MOD;
  return res;
}
int32_t main() {
  long long n, K, L;
  cin >> n >> K >> L;
  vector<long long> p;
  for (long long i = 0; i < n * K; i++) {
    long long x;
    cin >> x;
    p.push_back(x);
  }
  sort((p).begin(), (p).end());
  long long hi = upper_bound((p).begin(), (p).end(), p[0] + L) - p.begin();
  if (hi < n) {
    cout << 0 << endl;
    return 0;
  }
  vector<bool> used(hi);
  long long tot = 0;
  for (long long i = 0; i < hi; i += K) {
    used[i] = true;
    tot++;
  }
  long long ptr = hi - 1;
  while (tot < n) {
    assert(ptr >= 0);
    if (!used[ptr]) used[ptr] = true, tot++;
    ptr--;
  }
  long long ans = 0;
  for (long long i = 0; i < hi; i++) {
    if (used[i]) ans += p[i];
  }
  cout << ans << endl;
}

```