#include <bits/stdc++.h>
using namespace std;
int const N = 5e5 + 10, oo = 1e9;
int mod = oo + 7;
long long const OO = 1e18;
multiset<int> st;
int main() {
  int n, k, l;
  scanf("%d%d%d", &n, &k, &l);
  for (int i = 0; i < n * k; i++) {
    int a;
    scanf("%d", &a);
    st.insert(a);
  }
  long long sum = 0;
  int mn = *st.begin();
  int mx = *(--st.upper_bound(mn + l));
  while (!st.empty()) {
    int ls = oo + 1;
    for (int i = 0; i < k - 1; i++) {
      ls = *(--st.end());
      st.erase(--st.end());
    }
    if (ls > mx) {
      if (mx > mn + l || st.find(mx) == st.end()) {
        puts("0");
        return 0;
      }
      sum += mx;
      st.erase(st.find(mx));
    } else {
      if (st.find(mx) == st.end()) {
        if (*(--st.end()) > mn + l) {
          puts("0");
          return 0;
        }
        sum += *(--st.end());
        st.erase(--st.end());
      } else {
        if (mx > mn + l) {
          puts("0");
          return 0;
        }
        sum += mx;
        st.erase(st.find(mx));
      }
    }
    if (st.empty()) break;
    mx = *(--st.upper_bound(mn + l));
  }
  printf("%lld\n", sum);
  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;
int const N = 5e5 + 10, oo = 1e9;
int mod = oo + 7;
long long const OO = 1e18;
multiset<int> st;
int main() {
  int n, k, l;
  scanf("%d%d%d", &n, &k, &l);
  for (int i = 0; i < n * k; i++) {
    int a;
    scanf("%d", &a);
    st.insert(a);
  }
  long long sum = 0;
  int mn = *st.begin();
  int mx = *(--st.upper_bound(mn + l));
  while (!st.empty()) {
    int ls = oo + 1;
    for (int i = 0; i < k - 1; i++) {
      ls = *(--st.end());
      st.erase(--st.end());
    }
    if (ls > mx) {
      if (mx > mn + l || st.find(mx) == st.end()) {
        puts("0");
        return 0;
      }
      sum += mx;
      st.erase(st.find(mx));
    } else {
      if (st.find(mx) == st.end()) {
        if (*(--st.end()) > mn + l) {
          puts("0");
          return 0;
        }
        sum += *(--st.end());
        st.erase(--st.end());
      } else {
        if (mx > mn + l) {
          puts("0");
          return 0;
        }
        sum += mx;
        st.erase(st.find(mx));
      }
    }
    if (st.empty()) break;
    mx = *(--st.upper_bound(mn + l));
  }
  printf("%lld\n", sum);
  return 0;
}

```