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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>

/*
#include <atcoder/all>
using namespace atcoder;
*/
#pragma GCC target("avx2")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
//#include <boost/multiprecision/cpp_int.hpp>
using namespace std;


using dou =long double;
string yes="yes";
string Yes="Yes";
string YES="YES";
string no="no";
string No="No";
string NO="NO";

template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }
typedef long long ll;
typedef pair<int,int> P;
typedef pair<ll,ll> PL;
const ll mod = 1000000007ll;
//const ll mod = 10000000000ll;
//const ll mod = 10000;

struct mint {
  ll x; // typedef long long ll;
  mint(ll x=0):x((x%mod+mod)%mod){}
  mint operator-() const { return mint(-x);}
  mint& operator+=(const mint a) {
    if ((x += a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator-=(const mint a) {
    if ((x += mod-a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator*=(const mint a) { (x *= a.x) %= mod; return *this;}
  mint operator+(const mint a) const { return mint(*this) += a;}
  mint operator-(const mint a) const { return mint(*this) -= a;}
  mint operator*(const mint a) const { return mint(*this) *= a;}
  mint pow(ll t) const {
    if (!t) return 1;
    mint a = pow(t>>1);
    a *= a;
    if (t&1) a *= *this;
    return a;
  }

  // for prime mod
  mint inv() const { return pow(mod-2);}
  mint& operator/=(const mint a) { return *this *= a.inv();}
  mint operator/(const mint a) const { return mint(*this) /= a;}
};
istream& operator>>(istream& is, const mint& a) { return is >> a.x;}
ostream& operator<<(ostream& os, const mint& a) { return os << a.x;}
#define rep(i, n)         for(ll i = 0; i < (ll)(n); i++)
#define brep(n)           for(int bit=0;bit<(1<<n);bit++)
#define bbrep(n)           for(int bbit=0;bbit<(1<<n);bbit++)
#define erep(i,container) for (auto &i : container)
#define itrep(i,container) for (auto i : container)
#define irep(i, n)        for(ll i = n-1; i >= (ll)0ll; i--)
#define rrep(i,m,n) for(ll i = m; i < (ll)(n); i++)
#define reprep(i,j,h,w) rep(i,h)rep(j,w)
#define all(x) (x).begin(),(x).end()
#define rall(x) (x).rbegin(),(x).rend()
#define VEC(type,name,n) std::vector<type> name(n);rep(i,n)std::cin >> name[i];
#define pb push_back
#define pf push_front
#define query int qq;std::cin >> qq;rep(qqq,qq)
#define lb lower_bound
#define ub upper_bound
#define fi first
#define se second
#define itn int
#define mp make_pair
//#define sum(a) accumulate(all(a),0ll)
#define keta fixed<<setprecision
#define vout(a) erep(qxqxqx,a)std::cout << qxqxqx << ' ';std::cout  << std::endl;
#define vvector(name,typ,m,n,a)vector<vector<typ> > name(m,vector<typ> (n,a))
//#define vvector(name,typ,m,n)vector<vector<typ> > name(m,vector<typ> (n))
#define vvvector(name,t,l,m,n,a) vector<vector<vector<t> > > name(l, vector<vector<t> >(m, vector<t>(n,a)));
#define vvvvector(name,t,k,l,m,n,a) vector<vector<vector<vector<t> > > > name(k,vector<vector<vector<t> > >(l, vector<vector<t> >(m, vector<t>(n,a)) ));
#define case std::cout <<"Case #" <<qqq+1<<":"
#define RES(a,i,j) a.resize(i);rep(ii,i)a[ii].resize(j);
#define RESRES(a,i,j,k) a.resize(i);rep(ii,i)a[ii].resize(j);reprep(ii,jj,i,j){dp[ii][jj].resize(k)};
#define res resize
#define as assign
#define ffor for(;;)
#define ppri(a,b) std::cout << a<<" "<<b << std::endl
#define pppri(a,b,c) std::cout << a<<" "<<b <<" "<< c<<std::endl
#define ppppri(a,b,c,d) std::cout << a<<" "<<b <<" "<< c<<' '<<d<<std::endl
#define aall(x,n) (x).begin(),(x).begin()+(n)
#define SUM(a) accumulate(all(a),0ll) 
#define stirng string
#define gin(a,b) int a,b;std::cin >> a>>b;a--;b--;
#define popcount __builtin_popcount
#define permu(a) next_permutation(all(a))
//#define grid_input(a,type) int h,w;std::cin >> h>>w;vvector(a,type,h,w,0);reprep(i,j,h,w)std::cin >> a[i][j];

//typedef long long T;
ll ceil(ll a,ll b){
    return ((a+b-1)/b);
}
const int INF = 2000000000;
//const ll INF64 =3223372036854775807ll;
//const ll INF64 = 9223372036854775807ll;
const ll INF64 = 9223372036854775ll;
const ll MOD = 1000000007ll;
//const ll MOD = 1000003ll;
const ll OD = 1000000000000007ll;
const dou pi=3.141592653589793;
long long modpow(long long a, long long n,ll MOD) { //累乗の余剰
    long long res = 1;
    while (n > 0) {
        if (n & 1) res = res * a % MOD;
        a = a * a % MOD;
        n >>= 1;
    }
    return res;
}


//メモ
//ゲーム(Grundy数とか)の復習をする
//群論の勉強をする？
//ドツボにハマったら頑張って今までの思考をリセットする
//学会のスライドを作る(水曜日まで)
//周期性の実験をする
//リスニング力をどうにかする
//GT学会の会費を払う
/*
#include <atcoder/all>
using namespace atcoder;
*/
/*
#include <atcoder/all>
using namespace atcoder;
*/

int main(){
    int n,c,k;
    std::cin >> n>>c>>k;
    VEC(int,a,n);
    sort(all(a));
    int nin=c,ti=0;
    int ans=0;
    rep(i,n){
        if(nin==c||a[i]-ti>k){
            ans++;
            nin=1;ti=a[i];
        }
        else nin++;
     //   ppri(nin,ti);
    }
    std::cout << ans << std::endl;
    
}