#include <bits/stdc++.h>
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast,no-stack-protector")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimize("unroll-loops")
using namespace std;
template <typename T>
T gcd(T a, T b) {
  while (b) {
    T t = a % b;
    a = b;
    b = t;
  }
  return a;
}
template <typename T>
T lcm(T a, T b) {
  return a / gcd(a, b) * b;
}
template <typename T>
T gcd(T a, T b, T c) {
  return gcd(gcd(a, b), c);
}
template <typename T>
bool isPrime(T x) {
  if (x <= 4 || x % 2 == 0 || x % 3 == 0) return x == 2 || x == 3;
  for (T i = 5; i * i <= x; i += 6)
    if (x % i == 0 || x % (i + 2) == 0) return 0;
  return 1;
}
template <typename T>
bool bitcheck(T p, T pos) {
  return (bool)(p & (1 << pos));
}
template <typename T>
T biton(T p, T pos) {
  return p = p | (1 << pos);
}
template <typename T>
T bitoff(T p, T pos) {
  return p = p & ~(1 << pos);
}
template <typename T>
long long int toint(T s) {
  long long int p;
  stringstream ss(s);
  ss >> p;
  return p;
}
template <typename T>
string tostring(T n) {
  stringstream ss;
  ss << n;
  return ss.str();
}
template <typename T>
T POW(T b, T p) {
  T Ans = 1;
  while (p) {
    if (p & 1) Ans = (Ans * b);
    b = (b * b);
    p >>= 1;
  }
  return Ans;
}
template <typename T>
T BigMod(T b, T p, T Mod) {
  T Ans = 1;
  while (p) {
    if (p & 1) Ans = (Ans * b) % Mod;
    b = (b * b) % Mod;
    p >>= 1;
  }
  return Ans;
}
template <typename T>
T ModInverse(T p, T Mod) {
  return BigMod(p, Mod - 2, Mod);
}
void deb(istream_iterator<string> it) {}
template <typename T, typename... Args>
void deb(istream_iterator<string> it, T a, Args... args) {
  cerr << *it << " = " << a << endl;
  deb(++it, args...);
}
const int fx[] = {+0, +0, +1, -1, -1, +1, -1, +1};
const int fy[] = {-1, +1, +0, +0, +1, +1, -1, -1};
const int mx = 1e5 + 10;
const int inf = 0x3f3f3f3f;
const int mod = 1e9 + 7;
const double pi = 2 * acos(0.0);
const double E = 2.71828182845904523536;
const double eps = 1e-9;
vector<long long int> v;
int main() {
  long long int n, k, l, no;
  cin >> n >> k >> l;
  for (int i = 1; i <= n * k; i++) {
    cin >> no;
    v.push_back(no);
  }
  sort(v.begin(), v.end());
  long long int cnt = 0, a = 0, b, ans = 0;
  for (int i = v.size() - 1; i >= 0; i--) {
    cnt++;
    if (v[i] - v[0] <= l && cnt >= k) {
      cnt -= k;
      ans += v[i];
      a++;
    }
  }
  if (a == n)
    cout << ans << '\n';
  else
    cout << "0" << '\n';
  return 0;
}

### Prompt

Generate 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>
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast,no-stack-protector")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimize("unroll-loops")
using namespace std;
template <typename T>
T gcd(T a, T b) {
  while (b) {
    T t = a % b;
    a = b;
    b = t;
  }
  return a;
}
template <typename T>
T lcm(T a, T b) {
  return a / gcd(a, b) * b;
}
template <typename T>
T gcd(T a, T b, T c) {
  return gcd(gcd(a, b), c);
}
template <typename T>
bool isPrime(T x) {
  if (x <= 4 || x % 2 == 0 || x % 3 == 0) return x == 2 || x == 3;
  for (T i = 5; i * i <= x; i += 6)
    if (x % i == 0 || x % (i + 2) == 0) return 0;
  return 1;
}
template <typename T>
bool bitcheck(T p, T pos) {
  return (bool)(p & (1 << pos));
}
template <typename T>
T biton(T p, T pos) {
  return p = p | (1 << pos);
}
template <typename T>
T bitoff(T p, T pos) {
  return p = p & ~(1 << pos);
}
template <typename T>
long long int toint(T s) {
  long long int p;
  stringstream ss(s);
  ss >> p;
  return p;
}
template <typename T>
string tostring(T n) {
  stringstream ss;
  ss << n;
  return ss.str();
}
template <typename T>
T POW(T b, T p) {
  T Ans = 1;
  while (p) {
    if (p & 1) Ans = (Ans * b);
    b = (b * b);
    p >>= 1;
  }
  return Ans;
}
template <typename T>
T BigMod(T b, T p, T Mod) {
  T Ans = 1;
  while (p) {
    if (p & 1) Ans = (Ans * b) % Mod;
    b = (b * b) % Mod;
    p >>= 1;
  }
  return Ans;
}
template <typename T>
T ModInverse(T p, T Mod) {
  return BigMod(p, Mod - 2, Mod);
}
void deb(istream_iterator<string> it) {}
template <typename T, typename... Args>
void deb(istream_iterator<string> it, T a, Args... args) {
  cerr << *it << " = " << a << endl;
  deb(++it, args...);
}
const int fx[] = {+0, +0, +1, -1, -1, +1, -1, +1};
const int fy[] = {-1, +1, +0, +0, +1, +1, -1, -1};
const int mx = 1e5 + 10;
const int inf = 0x3f3f3f3f;
const int mod = 1e9 + 7;
const double pi = 2 * acos(0.0);
const double E = 2.71828182845904523536;
const double eps = 1e-9;
vector<long long int> v;
int main() {
  long long int n, k, l, no;
  cin >> n >> k >> l;
  for (int i = 1; i <= n * k; i++) {
    cin >> no;
    v.push_back(no);
  }
  sort(v.begin(), v.end());
  long long int cnt = 0, a = 0, b, ans = 0;
  for (int i = v.size() - 1; i >= 0; i--) {
    cnt++;
    if (v[i] - v[0] <= l && cnt >= k) {
      cnt -= k;
      ans += v[i];
      a++;
    }
  }
  if (a == n)
    cout << ans << '\n';
  else
    cout << "0" << '\n';
  return 0;
}

```