#include <bits/stdc++.h>
using namespace std;
inline char read() {
  static const int IN_LEN = 1000000;
  static char buf[IN_LEN], *s, *t;
  return (s == t               ? t = (s = buf) + fread(buf, 1, IN_LEN, stdin),
          (s == t ? -1 : *s++) : *s++);
}
template <class T>
inline void read(T &x) {
  static bool iosig;
  static char c;
  for (iosig = false, c = read(); !isdigit(c); c = read()) {
    if (c == '-') iosig = true;
    if (c == -1) return;
  }
  for (x = 0; isdigit(c); c = read()) x = ((x + (x << 2)) << 1) + (c ^ '0');
  if (iosig) x = -x;
}
const int OUT_LEN = 10000000;
char obuf[OUT_LEN], *ooh = obuf;
inline void print(char c) {
  if (ooh == obuf + OUT_LEN) fwrite(obuf, 1, OUT_LEN, stdout), ooh = obuf;
  *ooh++ = c;
}
template <class T>
inline void print(T x) {
  static int buf[30], cnt;
  if (x == 0)
    print('0');
  else {
    if (x < 0) print('-'), x = -x;
    for (cnt = 0; x; x /= 10) buf[++cnt] = x % 10 + 48;
    while (cnt) print((char)buf[cnt--]);
  }
}
inline void flush() { fwrite(obuf, 1, ooh - obuf, stdout); }
const int N = 100005;
int n, k, l, m, a[N];
bool vis[N];
long long ans;
int main() {
  read(n), read(k), read(l);
  m = n * k;
  for (register int i = (1); i <= (m); ++i) read(a[i]);
  sort(a + 1, a + m + 1);
  int j = 1;
  while (a[j + 1] - a[1] <= l && j <= m) ++j;
  if (j < n) return puts("0"), 0;
  int tot = n;
  for (register int i = (0); i < (n); ++i)
    if (i * k + 1 <= j && tot)
      vis[i * k + 1] = 1, --tot, ans += a[i * k + 1];
    else
      break;
  for (register int i = (j); i >= (1); --i)
    if (!vis[i] && tot) vis[i] = 1, --tot, ans += a[i];
  printf("%lld", ans);
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;
inline char read() {
  static const int IN_LEN = 1000000;
  static char buf[IN_LEN], *s, *t;
  return (s == t               ? t = (s = buf) + fread(buf, 1, IN_LEN, stdin),
          (s == t ? -1 : *s++) : *s++);
}
template <class T>
inline void read(T &x) {
  static bool iosig;
  static char c;
  for (iosig = false, c = read(); !isdigit(c); c = read()) {
    if (c == '-') iosig = true;
    if (c == -1) return;
  }
  for (x = 0; isdigit(c); c = read()) x = ((x + (x << 2)) << 1) + (c ^ '0');
  if (iosig) x = -x;
}
const int OUT_LEN = 10000000;
char obuf[OUT_LEN], *ooh = obuf;
inline void print(char c) {
  if (ooh == obuf + OUT_LEN) fwrite(obuf, 1, OUT_LEN, stdout), ooh = obuf;
  *ooh++ = c;
}
template <class T>
inline void print(T x) {
  static int buf[30], cnt;
  if (x == 0)
    print('0');
  else {
    if (x < 0) print('-'), x = -x;
    for (cnt = 0; x; x /= 10) buf[++cnt] = x % 10 + 48;
    while (cnt) print((char)buf[cnt--]);
  }
}
inline void flush() { fwrite(obuf, 1, ooh - obuf, stdout); }
const int N = 100005;
int n, k, l, m, a[N];
bool vis[N];
long long ans;
int main() {
  read(n), read(k), read(l);
  m = n * k;
  for (register int i = (1); i <= (m); ++i) read(a[i]);
  sort(a + 1, a + m + 1);
  int j = 1;
  while (a[j + 1] - a[1] <= l && j <= m) ++j;
  if (j < n) return puts("0"), 0;
  int tot = n;
  for (register int i = (0); i < (n); ++i)
    if (i * k + 1 <= j && tot)
      vis[i * k + 1] = 1, --tot, ans += a[i * k + 1];
    else
      break;
  for (register int i = (j); i >= (1); --i)
    if (!vis[i] && tot) vis[i] = 1, --tot, ans += a[i];
  printf("%lld", ans);
  return 0;
}

```