#include <bits/stdc++.h>
using namespace std;
const double pi = acos(-1.0);
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 << 1) + (x << 3) + ch - '0';
    ch = getchar();
  }
  return f * x;
}
inline void write(long long a) {
  if (a < 0) putchar('-'), a = -a;
  if (a >= 10) write(a / 10), putchar(a % 10 + '0');
}
const int MAX_N = 1e5 + 10;
long long a[MAX_N], n, m, k;
int main() {
  scanf("%lld%lld%lld", &n, &k, &m);
  for (int i = 0; i < n * k; i++) {
    scanf("%lld", a + i);
  }
  sort(a, a + n * k);
  int t = upper_bound(a, a + n * k, a[0] + m) - a;
  if (t >= n) {
    long long ans = 0;
    int tt = t;
    int pos = 0;
    int cnt = 0;
    for (int i = k; i >= 1 && tt; i--) {
      while (tt - i + cnt + 1 >= n && tt >= i) {
        tt -= i;
        ans += a[pos];
        pos += i;
        cnt++;
      }
      if (tt + cnt == n) {
        break;
      }
    }
    for (int i = pos; i < t; i++) {
      ans += a[i];
    }
    printf("%lld\n", ans);
  } else {
    puts("0");
  }
  return 0;
}

### Prompt

Generate 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 double pi = acos(-1.0);
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 << 1) + (x << 3) + ch - '0';
    ch = getchar();
  }
  return f * x;
}
inline void write(long long a) {
  if (a < 0) putchar('-'), a = -a;
  if (a >= 10) write(a / 10), putchar(a % 10 + '0');
}
const int MAX_N = 1e5 + 10;
long long a[MAX_N], n, m, k;
int main() {
  scanf("%lld%lld%lld", &n, &k, &m);
  for (int i = 0; i < n * k; i++) {
    scanf("%lld", a + i);
  }
  sort(a, a + n * k);
  int t = upper_bound(a, a + n * k, a[0] + m) - a;
  if (t >= n) {
    long long ans = 0;
    int tt = t;
    int pos = 0;
    int cnt = 0;
    for (int i = k; i >= 1 && tt; i--) {
      while (tt - i + cnt + 1 >= n && tt >= i) {
        tt -= i;
        ans += a[pos];
        pos += i;
        cnt++;
      }
      if (tt + cnt == n) {
        break;
      }
    }
    for (int i = pos; i < t; i++) {
      ans += a[i];
    }
    printf("%lld\n", ans);
  } else {
    puts("0");
  }
  return 0;
}

```