Problem:
Score : 300 points

Problem Statement

Every day, N passengers arrive at Takahashi Airport.
The i-th passenger arrives at time T_i.

Every passenger arrived at Takahashi airport travels to the city by bus. Each bus can accommodate up to C passengers.
Naturally, a passenger cannot take a bus that departs earlier than the airplane arrives at the airport.
Also, a passenger will get angry if he/she is still unable to take a bus K units of time after the arrival of the airplane.
For that reason, it is necessary to arrange buses so that the i-th passenger can take a bus departing at time between T_i and T_i + K (inclusive).

When setting the departure times for buses under this condition, find the minimum required number of buses.
Here, the departure time for each bus does not need to be an integer, and there may be multiple buses that depart at the same time.

Constraints

2 \leq N \leq 100000

1 \leq C \leq 10^9

1 \leq K \leq 10^9

1 \leq T_i \leq 10^9

C, K and T_i are integers.

Input

The input is given from Standard Input in the following format:

N C K
T_1
T_2
:
T_N

Output

Print the minimum required number of buses.

Sample Input 1

5 3 5
1
2
3
6
12

Sample Output 1

3

For example, the following three buses are enough:

A bus departing at time 4.5, that carries the passengers arriving at time 2 and 3.

A bus departing at time 6, that carries the passengers arriving at time 1 and 6.

A bus departing at time 12, that carries the passenger arriving at time 12.

Sample Input 2

6 3 3
7
6
2
8
10
6

Sample Output 2

3

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;

//conversion
//------------------------------------------
inline int toInt(string s) {int v; istringstream sin(s);sin>>v;return v;}
template<class T> inline string toString(T x) {ostringstream sout;sout<<x;return sout.str();}

//math
//-------------------------------------------
template<class T> inline T sqr(T x) {return x*x;}

//typedef
//------------------------------------------
typedef long long LL;
typedef pair<int, int> PII;
typedef pair<LL,LL> PLL;
typedef vector<int> VI;
typedef vector<VI> VVI;
typedef vector<LL> VLL;
typedef vector<VLL> VVLL;
typedef vector<bool> VB;
typedef vector<VB> VVB;
typedef vector<double> VD;
typedef vector<VD> VVD;
typedef vector<string> VS;
typedef vector<VS> VVS;
typedef vector<char> VC;
typedef vector<VC> VVC;
typedef vector<PII> VPII;
typedef vector<PLL> VPLL;
typedef priority_queue<int> PQGI; //大きい順
typedef priority_queue<int, VI, greater<int>> PQLI;
typedef priority_queue<PII> PQGP;
typedef priority_queue<PII, VPII, greater<int>> PQLP;

//container util
//------------------------------------------
#define ALL(a)  (a).begin(),(a).end()
#define RALL(a) (a).rbegin(),(a).rend()
#define EB emplace_back
#define PB push_back
#define PF push_front
#define POB pop_back
#define POF pop_front
#define MP make_pair
#define SZ(a) int((a).size())
#define SQ(a) ((a)*(a))
#define EACH(i,c) for(typeof((c).begin()) i=(c).begin(); i!=(c).end(); ++i)
#define EXIST(s,e) ((s).find(e)!=(s).end())
#define SORT(c) sort((c).begin(),(c).end())
#define SORTR(c) sort((c).rbegin(), (c).rend())
#define LB lower_bound
#define UB upper_bound
#define NEXP next_permutation
#define FI first
#define SE second
#define Vmin(a) *min_element((a).begin(),(a).end())
#define Vmax(a) *max_element((a).begin(),(a).end())

//repetition
//------------------------------------------
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n)  FOR(i,0,n)
#define FORR(i,a,b) for(int i = (b-1);i>=a;i--)
#define REPR(i,n) FORR(i,0,n)
#define BFOR(bit,a,b) for(int bit = (a); bit < (1<<(b)); ++bit)
#define BREP(bit,n) BFOR(bit,0,n)

//constant
//--------------------------------------------
const double EPS = 1e-10;
const double PI  = acos(-1.0);
const int INF = INT_MAX/2;
const LL LINF = LLONG_MAX/3;
const int RINF = INT_MIN/2;
const LL RLINF = LLONG_MIN/3;
const LL MOD = 1e9+7;
const LL MODD = 998244353;
const int MAX = 510000;

inline bool Eq(double a, double b) { return fabs(b - a) < EPS; }

//other
//-------------------------------------------
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }

#define COUT(x) cout << (x) << endl
#define COUT2(x,y) cout << (x) << " " << (y) << endl
#define COUT3(x,y,z) cout << (x) << " " << (y) << " " << (z) << endl
#define PR(x) cout << (x)
#define ENDL cout << endl
#define SPACE cout << " "
#define BC(x) __builtin_popcountll(x)


VI dx = {1,0,-1,0,1,1,-1,-1};
VI dy = {0,1,0,-1,1,-1,1,-1};
VC dc = {'R','D','L','U'};
VS ds = {"R","D","L","U","RD","RU","LD","LU"};

LL Gcd(LL a, LL b) {return __gcd(a,b);}
LL Lcm(LL a, LL b) {return a/Gcd(a,b)*b;}

LL ModPow(LL A, LL N, LL M){
  LL res = 1;
  while(N>0){
    if(N&1) res = res * A % M;
    A = A * A % M;
    N >>= 1;
  }
  return res;
}


template<class T> void VCout(vector<T> &v){
  int N = SZ(v);
  REP(i,N){
    cout << v.at(i);
    if(i == N-1) cout << endl;
    else cout << " ";
  }
  return;
}



int main() {

  // cin.tie(0);
  // ios::sync_with_stdio(false);
  cout << fixed << setprecision(12);

  int n;
  LL c,k; cin >> n >> c >> k;
  VLL t(n);
  REP(i,n) cin >> t.at(i);
  SORT(t);

  LL m = 0;
  LL time = 0;
  int ans = 0;

  REP(i,n){
    time = t.at(i)+k;
    ans++;
    m=1;
    while(i+1<n){
      if(m+1<=c && time >= t.at(i+1)){
        i++;
        m++;
      }else{
        break;
      }
    }
  }
  
  COUT(ans);

  return 0;
}