#include <bits/stdc++.h>
using namespace std;
const int inf = 1 << 30;
const int N = 100100;
int n, a[N], k, l;
int main() {
  scanf("%d %d %d", &n, &k, &l);
  for (int i = 1; i <= n * k; i++) scanf("%d", a + i);
  int nn = n * k;
  sort(a + 1, a + 1 + nn);
  int idx = 1;
  while (idx <= nn) {
    if (a[idx] - a[1] <= l)
      idx++;
    else
      break;
  }
  if (idx > nn || a[idx] - a[1] > l) idx--;
  if (idx < n) {
    printf("0");
    return 0;
  }
  vector<int> v, vv;
  for (int j = idx + 1; j <= nn; j++) v.push_back(a[j]);
  for (int i = 1; i <= idx; i++) vv.push_back(a[i]);
  long long ANS = 0;
  for (int i = idx - 1; i >= 0; --i) {
    if (v.size() < k - 1) break;
    int cnt = k - 1;
    while (cnt--) v.pop_back();
    ANS += vv.back();
    vv.pop_back();
  }
  for (auto x : v) vv.push_back(x);
  for (int i = 0; i < vv.size(); i += k) {
    ANS += vv[i];
  }
  printf("%lld", ANS);
  return 0;
}

### Prompt

Your task is to create 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 inf = 1 << 30;
const int N = 100100;
int n, a[N], k, l;
int main() {
  scanf("%d %d %d", &n, &k, &l);
  for (int i = 1; i <= n * k; i++) scanf("%d", a + i);
  int nn = n * k;
  sort(a + 1, a + 1 + nn);
  int idx = 1;
  while (idx <= nn) {
    if (a[idx] - a[1] <= l)
      idx++;
    else
      break;
  }
  if (idx > nn || a[idx] - a[1] > l) idx--;
  if (idx < n) {
    printf("0");
    return 0;
  }
  vector<int> v, vv;
  for (int j = idx + 1; j <= nn; j++) v.push_back(a[j]);
  for (int i = 1; i <= idx; i++) vv.push_back(a[i]);
  long long ANS = 0;
  for (int i = idx - 1; i >= 0; --i) {
    if (v.size() < k - 1) break;
    int cnt = k - 1;
    while (cnt--) v.pop_back();
    ANS += vv.back();
    vv.pop_back();
  }
  for (auto x : v) vv.push_back(x);
  for (int i = 0; i < vv.size(); i += k) {
    ANS += vv[i];
  }
  printf("%lld", ANS);
  return 0;
}

```