#include <bits/stdc++.h>
using namespace std;
struct input {
  long long int p, c;
};
const long long int M = 1e9 + 7;
const long long int N = 1e5;
template <class T>
T gcd(const T a, const T b) {
  return (b ? gcd<T>(b, a % b) : a);
}
bool compare(input a, input b) { return a.p < b.p; }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int l, n, m, i, j, k, temp, x, y, z, t, p, mx, mn;
  string s;
  cin >> n >> k >> l;
  long long int sz = n * k;
  input a[sz + 10];
  for (i = 1; i <= sz; i++) {
    cin >> a[i].p;
    a[i].c = -1;
  }
  sort(a + 1, a + sz + 1, compare);
  if (a[n].p - a[1].p > l) {
    cout << 0 << endl;
    return 0;
  }
  long long int lcur, rcur;
  for (i = 1; i <= sz; i++) {
    if (a[i].p - a[1].p <= l) lcur = i;
  }
  lcur = min(n * k - k + 1, lcur);
  rcur = n * k;
  long long int pos, cnt = 0, ct = 0, sum = 0;
  while (cnt != n) {
    ct = 0;
    while (ct != k - 1) {
      if (a[rcur].c == -1) ct++;
      rcur--;
    }
    pos = min(lcur, rcur);
    lcur = pos - 1;
    sum += a[pos].p;
    a[pos].c = 1;
    cnt++;
  }
  cout << sum << endl;
  return 0;
}

### Prompt

In cpp, your task is to solve 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;
struct input {
  long long int p, c;
};
const long long int M = 1e9 + 7;
const long long int N = 1e5;
template <class T>
T gcd(const T a, const T b) {
  return (b ? gcd<T>(b, a % b) : a);
}
bool compare(input a, input b) { return a.p < b.p; }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int l, n, m, i, j, k, temp, x, y, z, t, p, mx, mn;
  string s;
  cin >> n >> k >> l;
  long long int sz = n * k;
  input a[sz + 10];
  for (i = 1; i <= sz; i++) {
    cin >> a[i].p;
    a[i].c = -1;
  }
  sort(a + 1, a + sz + 1, compare);
  if (a[n].p - a[1].p > l) {
    cout << 0 << endl;
    return 0;
  }
  long long int lcur, rcur;
  for (i = 1; i <= sz; i++) {
    if (a[i].p - a[1].p <= l) lcur = i;
  }
  lcur = min(n * k - k + 1, lcur);
  rcur = n * k;
  long long int pos, cnt = 0, ct = 0, sum = 0;
  while (cnt != n) {
    ct = 0;
    while (ct != k - 1) {
      if (a[rcur].c == -1) ct++;
      rcur--;
    }
    pos = min(lcur, rcur);
    lcur = pos - 1;
    sum += a[pos].p;
    a[pos].c = 1;
    cnt++;
  }
  cout << sum << endl;
  return 0;
}

```