#include <bits/stdc++.h>
using namespace std;
const int N = 100005;
inline long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
long long n, K, m, tot, a[N];
int main() {
  n = read();
  K = read();
  m = read();
  tot = n * K;
  for (long long i = 1; i <= tot; i++) a[i] = read();
  sort(a + 1, a + tot + 1);
  long long r = 1, cnt = 1;
  while (r < tot && a[r + 1] - a[1] <= m) r++;
  if (r < n) {
    puts("0");
    return 0;
  }
  cnt = r - n;
  long long k1 = 0, k2 = 0;
  if (K > 1) k1 = cnt / (K - 1), k2 = cnt % (K - 1);
  long long ans = 0;
  for (long long i = 0; i < k1; i++) {
    ans += a[i * K + 1];
  }
  if (k2) ans += a[k1 * K + 1];
  long long st = k1 * K + k2;
  if (k2)
    st += 2;
  else
    st++;
  for (long long i = st; i <= r; i++) ans += a[i];
  cout << ans << endl;
  return 0;
}

### Prompt

Your challenge is to write 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 int N = 100005;
inline long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
long long n, K, m, tot, a[N];
int main() {
  n = read();
  K = read();
  m = read();
  tot = n * K;
  for (long long i = 1; i <= tot; i++) a[i] = read();
  sort(a + 1, a + tot + 1);
  long long r = 1, cnt = 1;
  while (r < tot && a[r + 1] - a[1] <= m) r++;
  if (r < n) {
    puts("0");
    return 0;
  }
  cnt = r - n;
  long long k1 = 0, k2 = 0;
  if (K > 1) k1 = cnt / (K - 1), k2 = cnt % (K - 1);
  long long ans = 0;
  for (long long i = 0; i < k1; i++) {
    ans += a[i * K + 1];
  }
  if (k2) ans += a[k1 * K + 1];
  long long st = k1 * K + k2;
  if (k2)
    st += 2;
  else
    st++;
  for (long long i = st; i <= r; i++) ans += a[i];
  cout << ans << endl;
  return 0;
}

```