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: Wrong Answer)

#include <iostream>
#include <cstring>
#include <map>
#include <iomanip>
#include <algorithm>
#include <cmath>
#include <set>
#include <vector>
#include <queue>
#include <list>
#include <numeric>
#include <stdio.h>
#include <string>
#include <cstdlib>
#include <math.h>
#include <stack>
#include <climits>
#include <bitset>
#include <utility>


using namespace std;
typedef long long ll;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }

//最大公約数
ll gcd(ll x, ll y) {
    ll tmp = 0;
    if (x < y){
        tmp=x;
        x=y;
        y=tmp;
    }
    while (y > 0) {
        ll r = x % y;
        x = y;
        y = r;
        }
    return x;
}

//最大公倍数
ll lcm(ll x,ll y){
    return x/gcd(x,y)*y;
}

const int MAX = 1e6+1;
const ll MOD = 1e9+7;

long long fac[MAX], finv[MAX], inv[MAX];

// テーブルを作る前処理
void COMinit() {
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (int i = 2; i < MAX; i++){
        fac[i] = fac[i - 1] * i % MOD;
        inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
        finv[i] = finv[i - 1] * inv[i] % MOD;
    }
}

// 二項係数計算
long long COM(int n, int k){
    if (n < k) return 0;
    if (n < 0 || k < 0) return 0;
    return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}


//階乗
ll kaijo(ll k){
    ll sum = 1;
    for (ll i = 1; i <= k; ++i)
    {
        sum *= i;
        sum%=1000000000+7;
    }
    return sum;
}

long long modpow(long long a, long long n, long long mod) {
    long long res = 1;
    while (n > 0) {
        if (n & 1) res = res * a % mod;
        a = a * a % mod;
        n >>= 1;
    }
    return res;
}

//sieve(MAX);でベクトルIsPrimeに「MAXまでの数値の素数の真偽」を格納する。
vector<bool> IsPrime;
void sieve(size_t max){
    if(max+1 > IsPrime.size()){     // resizeで要素数が減らないように
        IsPrime.resize(max+1,true); // IsPrimeに必要な要素数を確保
    } 
    IsPrime[0] = false; // 0は素数ではない
    IsPrime[1] = false; // 1は素数ではない

    for(size_t i=2; i*i<=max; ++i) // 0からsqrt(max)まで調べる
        if(IsPrime[i]) // iが素数ならば
            for(size_t j=2; i*j<=max; ++j) // (max以下の)iの倍数は
                IsPrime[i*j] = false;      // 素数ではない
}

struct UnionFind {
    vector<int> par;
    UnionFind(int n) : par(n, -1) { }
    int root(int x) {
        if (par[x] < 0) return x;
        else return par[x] = root(par[x]);
    }
    bool issame(int x, int y) {
        return root(x) == root(y);
    }
    bool merge(int x, int y) {
        x = root(x); y = root(y);
        if (x == y) return false;
        if (par[x] > par[y]) swap(x, y); // merge technique
        par[x] += par[y];
        par[y] = x;
        return true;
    }
    int size(int x) {
        return -par[root(x)];
    }
};

ll count(int n, int a){
    ll bunshi = 1;
    for(int i = 0; i < a; i++){
        bunshi*=(n-i);
        bunshi%=MOD;
    }
    ll bunbo = 1;
    for(int i = 1; i < a+1; i++){
        bunbo*=i;
        bunbo%=MOD;
    }
    bunbo = modpow(bunbo,MOD-2,MOD);
    // cout<<(bunshi*bunbo)%MOD<<endl;
    return (bunshi*bunbo)%MOD;
}

// 約数列挙。約数をvector<ll>で返す。計算量はsqrt(n)。
vector<ll> divisor(ll n) {
    vector<ll> ret;
    for(ll i = 1; i * i <= n; i++) {
        if(n % i == 0) {
        ret.push_back(i);
        if(i * i != n) ret.push_back(n / i);
        }
    }
    sort(begin(ret), end(ret));
    return (ret);
}

// ここから開始
int main(){
    ll N,C,K;
    cin>>N>>C>>K;
    ll t[N];
    for(int i = 0; i < N; i++){
        cin>>t[i];
    }
    sort(t,t+N);
    ll i = 0;
    ll start;
    ll ans = 0;
    while(i<N){
        start = t[i];
        ll capa = 0;
        while(t[i]<=start+K&&capa<=C){
            i++;
            capa++;
        }
        ans++;
    }
    cout<<ans<<endl;
    return 0;
}