#include <bits/stdc++.h>
using namespace std;
vector<vector<int>> g;
vector<int> parent;
vector<int> dp, depth, vrtcnt;
int max_depth = 0, cur_depth = -1;
void dfs(int x, int p) {
  if (g[x].size() == 0) {
    depth[x] = 0;
    dp[x] = 0;
    if (p != -1) depth[p] = 1;
  }
}
int main() {
  int n, k, l;
  cin >> n >> k >> l;
  int m = n * k;
  vector<int> v(m, 0), used(m, 0);
  for (int i = 0; i < m; i++) cin >> v[i];
  sort(v.begin(), v.end());
  if (v[n - 1] - v[0] > l) {
    cout << 0;
    return 0;
  }
  int i = 1;
  while (i < m && v[i] - v[0] <= l) i++;
  i--;
  long long sum = v[0];
  int cnt = 1, dist = 0;
  for (int j = 1; j <= i; j++) {
    dist++;
    if (dist == k) {
      cnt++;
      sum += v[j];
      dist = 0;
      used[j] = 1;
    }
  }
  int j = i;
  while (cnt < n) {
    if (used[j] == 0) {
      cnt++;
      sum += v[j];
    }
    j--;
  }
  cout << sum;
  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;
vector<vector<int>> g;
vector<int> parent;
vector<int> dp, depth, vrtcnt;
int max_depth = 0, cur_depth = -1;
void dfs(int x, int p) {
  if (g[x].size() == 0) {
    depth[x] = 0;
    dp[x] = 0;
    if (p != -1) depth[p] = 1;
  }
}
int main() {
  int n, k, l;
  cin >> n >> k >> l;
  int m = n * k;
  vector<int> v(m, 0), used(m, 0);
  for (int i = 0; i < m; i++) cin >> v[i];
  sort(v.begin(), v.end());
  if (v[n - 1] - v[0] > l) {
    cout << 0;
    return 0;
  }
  int i = 1;
  while (i < m && v[i] - v[0] <= l) i++;
  i--;
  long long sum = v[0];
  int cnt = 1, dist = 0;
  for (int j = 1; j <= i; j++) {
    dist++;
    if (dist == k) {
      cnt++;
      sum += v[j];
      dist = 0;
      used[j] = 1;
    }
  }
  int j = i;
  while (cnt < n) {
    if (used[j] == 0) {
      cnt++;
      sum += v[j];
    }
    j--;
  }
  cout << sum;
  return 0;
}

```