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 <iostream>
#include <cstdio>
#include <string>
#include <vector>
#include <algorithm>
#include <functional>
#include <iomanip>
#include <stdlib.h>
#include <string.h>
#include <cstring>
#include <cmath>
#include <map>
#include <queue>
#include <deque>
#include <stack>
#include <set>
#include <stdio.h>
#include <tuple>
#include <bitset>
#include <cfloat>
#include <fstream>
#include <limits.h>
#include <list>
#include <math.h>
#include <random>
#include <unordered_map>
#include <unordered_set>
#include <bitset>
#include <numeric>
#include <utility>
#include <sstream>
#include <cassert>
template<class T> inline bool chmin(T& a, T b) {
	if (a > b) {
		a = b;
		return true;
	}
	return false;
}
template<class T> inline bool chmax(T& a, T b) {
	if (a < b) {
		a = b;
		return true;
	}
	return false;
}
using namespace std;
#define int long long
#define ll long long
#define rep(i, n) for (ll i = 0; i < (n); i++)
#define P pair<ll, ll>
#define sz(x) (ll)x.size()
#define ALL(x) (x).begin(),(x).end()
#define ALLR(x) (x).rbegin(),(x).rend()
#define VE vector<ll>
#define COUT(x) cout<<(x)<<endl
#define MA map<ll,ll>
#define SE set<ll>
#define PQ priority_queue<ll>
#define PQR priority_queue<ll,VE,greater<ll>>
#define YES(n) cout << ((n) ? "YES" : "NO"  ) << endl
#define Yes(n) cout << ((n) ? "Yes" : "No"  ) << endl
#define EPS (1e-14)
#define pb push_back
long long MOD = 1000000007;
//const long long MOD = 998244353;
const long long INF = 1LL << 60;
const double PI = acos(-1.0);
using Graph = vector<VE>;
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 {
		mint res(*this);
		return res += a;
	}
	mint operator-(const mint a) const {
		mint res(*this);
		return res -= a;
	}
	mint operator*(const mint a) const {
		mint res(*this);
		return res *= 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 {
		mint res(*this);
		return res /= a;
	}
	friend ostream& operator<<(ostream& os, const mint& m) {
		os << m.x;
		return os;
	}
};
struct combination {
	vector<mint> fact, ifact;
	combination(int n) :fact(n + 1), ifact(n + 1) {
		//assert(n < MOD);
		fact[0] = 1;
		for (int i = 1; i <= n; ++i) fact[i] = fact[i - 1] * i;
		ifact[n] = fact[n].inv();
		for (int i = n; i >= 1; --i) ifact[i - 1] = ifact[i] * i;
	}
	mint operator()(int n, int k) {
		if (k < 0 || k > n) return 0;
		return fact[n] * ifact[k] * ifact[n - k];
	}
}com(10);
struct Sieve {
	int n;
	vector<int> f, primes;
	
	Sieve(int n = 1) :n(n), f(n + 1) {
		f[0] = f[1] = -1;
		for (ll i = 2; i <= n; ++i) {
			if (f[i]) continue;
			primes.push_back(i);
			f[i] = i;
			for (ll j = i * i; j <= n; j += i) {
				if (!f[j]) f[j] = i;
			}
		}
	}
	
	bool isPrime(int x) { return f[x] == x; }
	
	vector<int> factorList(int x) {
		vector<int> res;
		while (x != 1) {
			res.push_back(f[x]);
			x /= f[x];
		}
		return res;
	}
	
	vector<P> factor(int x) {
		vector<int> fl = factorList(x);
		if (fl.size() == 0) return {};
		vector<P> res(1, P(fl[0], 0));
		for (int p : fl) {
			if (res.back().first == p) {
				res.back().second++;
			}
			else {
				res.emplace_back(p, 1);
			}
		}
		return res;
	}
};
class UnionFind {
public:
	vector <ll> par; 
	vector <ll> siz; 

	// Constructor
	UnionFind(ll sz_) : par(sz_), siz(sz_, 1) {
		for (ll i = 0; i < sz_; ++i) par[i] = i; 
	}
	void init(ll sz_) {
		par.resize(sz_);
		siz.resize(sz_, 1);
		for (ll i = 0; i < sz_; ++i) par[i] = i; 
	}

	// Member Function
	// Find
	ll root(ll x) { 
		while (par[x] != x) {
			x = par[x] = par[par[x]]; 
		}
		return x;
	}

	// Union(Unite, Merge)
	bool merge(ll x, ll y) {
		x = root(x);
		y = root(y);
		if (x == y) return false;
		
		if (siz[x] < siz[y]) swap(x, y);
		siz[x] += siz[y];
		par[y] = x;
		return true;
	}

	bool issame(ll x, ll y) {
		return root(x) == root(y);
	}

	ll size(ll x) {
		return siz[root(x)];
	}
};
template<class t> t gcd(t a, t b) { return b != 0 ? gcd(b, a % b) : a; }
ll lcm(ll a, ll b) {
	ll g = gcd(a, b);
	return a * b / g;
}
bool prime(ll n) {
	for (ll i = 2; i <= sqrt(n); i++) {
		if (n%i == 0)return false;
	}
	return n != 1;
}
map<ll, ll> prime_factor(ll n) {
	map<ll, ll> ret;
	for (ll i = 2; i * i <= n; i++) {
		while (n % i == 0) {
			ret[i]++;
			n /= i;
		}
	}
	if (n != 1) ret[n] = 1;
	return ret;
}
ll modinv(ll a, ll m) {
	ll b = m, u = 1, v = 0;
	while (b) {
		ll t = a / b;
		a -= t * b; swap(a, b);
		u -= t * v; swap(u, v);
	}
	u %= m;
	if (u < 0) u += m;
	return u;
}
vector<pair<char, int>> RunLength(string s) {
	if (s.size() == 0)return {};
	vector<pair<char, int>>res(1, pair<char, int>(s[0], 0));
	for (char p : s) {
		if (res.back().first == p) {
			res.back().second++;
		}
		else {
			res.emplace_back(p, 1);
		}
	}
	return res;
}
// Digit Count
int GetDigit(int num) {
	return log10(num) + 1;
}
// bit calculation[how many "1"] (= __builtin_popcount())
int bit_count(int n) {
	int cnt = 0;
	while (n > 0) {
		if (n % 2 == 1)cnt++;
		n /= 2;
	}
	return cnt;
}
mint POW(mint n, int p) {
	if (p == 0)return 1;
	if (p % 2 == 0) {
		mint t = POW(n, p / 2);
		return t * t;
	}
	return n * POW(n, p - 1);
}


const int dx[4] = { 1,0,-1,0 };
const int dy[4] = { 0,1,0,-1 };

struct edge { ll to, cost; };
typedef long double ld;



signed main() {
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	//cout << fixed << setprecision(15);
	int n, c, k; cin >> n >> c >> k;
	VE a(n); rep(i, n)cin >> a[i];
	sort(ALL(a));
	int ans = 0;
	int cnt = 0;
	int now = 0;
	rep(i, n) {
		if (cnt >= c || a[i] > now) {
			ans++; cnt = 0; now = a[i] + k;
		}
		cnt++;
	}
	cout << ans << endl;
	return 0;

}