#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:336777216")
using namespace std;
const double pi = atan((double)1.0) * 4;
long long MOD = (long long)1e9 + 7;
long long power(long long n, long long k) {
  if (k == 0)
    return 1;
  else if (k == 1)
    return n;
  else {
    if (k % 2 == 0) {
      return power((n * n) % MOD, k / 2);
    } else {
      return (power((n * n) % MOD, k / 2) * n) % MOD;
    }
  }
}
long long modInv(long long n) { return power(n, MOD - 2); }
long long gcd(long long a, long long b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
template <class T>
bool reverse_sort(T a, T b) {
  return a > b;
}
multiset<long long> arr, small, large;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n, k, l;
  cin >> n >> k >> l;
  long long low = LLONG_MAX;
  long long high = 0;
  for (int i = 0; i < n * k; i++) {
    long long t;
    cin >> t;
    arr.insert(t);
    low = min(low, t);
    high = max(high, t);
  }
  long long ok = low + l;
  for (auto it = arr.begin(); it != arr.end(); it++) {
    if (*it <= ok) {
      small.insert(-*it);
    } else {
      large.insert(*it);
    }
  }
  bool isok = true;
  long long ans = 0;
  for (auto it = large.begin(); it != large.end();) {
    long long used = 0;
    long long value = *it;
    while (it != large.end() && used < k - 1) {
      used++;
      large.erase(it++);
    }
    auto jt = small.begin();
    while (jt != small.end() && used < k) {
      used++;
      value = min(value, -*jt);
      small.erase(jt++);
    }
    if (used < k) {
      isok = false;
      break;
    }
    ans += value;
  }
  if (isok == false) {
    cout << 0;
  } else {
    long long value = LLONG_MAX;
    long long used = 0;
    for (auto it = small.begin(); it != small.end();) {
      value = min(value, -*it);
      used++;
      small.erase(it++);
      if (used == k) {
        ans += value;
        value = LLONG_MAX;
        used = 0;
      }
    }
    cout << ans;
  }
  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>
#pragma comment(linker, "/STACK:336777216")
using namespace std;
const double pi = atan((double)1.0) * 4;
long long MOD = (long long)1e9 + 7;
long long power(long long n, long long k) {
  if (k == 0)
    return 1;
  else if (k == 1)
    return n;
  else {
    if (k % 2 == 0) {
      return power((n * n) % MOD, k / 2);
    } else {
      return (power((n * n) % MOD, k / 2) * n) % MOD;
    }
  }
}
long long modInv(long long n) { return power(n, MOD - 2); }
long long gcd(long long a, long long b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
template <class T>
bool reverse_sort(T a, T b) {
  return a > b;
}
multiset<long long> arr, small, large;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n, k, l;
  cin >> n >> k >> l;
  long long low = LLONG_MAX;
  long long high = 0;
  for (int i = 0; i < n * k; i++) {
    long long t;
    cin >> t;
    arr.insert(t);
    low = min(low, t);
    high = max(high, t);
  }
  long long ok = low + l;
  for (auto it = arr.begin(); it != arr.end(); it++) {
    if (*it <= ok) {
      small.insert(-*it);
    } else {
      large.insert(*it);
    }
  }
  bool isok = true;
  long long ans = 0;
  for (auto it = large.begin(); it != large.end();) {
    long long used = 0;
    long long value = *it;
    while (it != large.end() && used < k - 1) {
      used++;
      large.erase(it++);
    }
    auto jt = small.begin();
    while (jt != small.end() && used < k) {
      used++;
      value = min(value, -*jt);
      small.erase(jt++);
    }
    if (used < k) {
      isok = false;
      break;
    }
    ans += value;
  }
  if (isok == false) {
    cout << 0;
  } else {
    long long value = LLONG_MAX;
    long long used = 0;
    for (auto it = small.begin(); it != small.end();) {
      value = min(value, -*it);
      used++;
      small.erase(it++);
      if (used == k) {
        ans += value;
        value = LLONG_MAX;
        used = 0;
      }
    }
    cout << ans;
  }
  return 0;
}

```