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)

#pragma GCC optimize("O3")
#include <bits/stdc++.h>
using namespace std;
typedef  long long ll;
double pi=3.1415926535898;
//double pi=acos(-1.0);
#define speed      ios::sync_with_stdio(false);cin.tie(NULL); cout.tie(NULL);
#define rng(i,a,b) for(int i=a;i>=int(b);i--)
#define rep(i,a,b) for(int i=a;i<=int(b);++i)
#define lop(i,n)   for(int i=0;i<int(n);i++)
#define ms(x,a)    memset(x,a,sizeof(x))
#define all(x)     (x).begin(),(x).end()
#define sz(x)      int(x.size())
#define pb         push_back
#define r0         return 0;
#define CC         continue;
#define bb(x)      cerr<< x <<endl;
#define pii        pair<int,int>
#define pis        pair<int,string>
#define vi         vector<int>
#define F          first
#define S          second
//#define max(a,b) a<b?b:a  __builtin_popcount;
//#define min(a,b) a<b?a:b
//freopen("output.txt","w",stdout),freopen("input.txt","r",stdin);
const ll M  = 1e9+7 ;
const int N = 50500;
const int inf = INT_MAX;
const double eps = 1e-7;
//////////////////////    Hi    ^_^  hack me if you can :P

ll n, c,k ;
ll t[100100];
ll lo, hi, md ;
bool ok (ll x)
{
    ll cnt = 0;
    ll cur = 0;
    ll mxt = 0;
    mxt = t[0] + k;
    cnt = cur = 1;
    for (int i=1; i<n; i++)
    {
        if (t[i]>mxt)
        {
            cnt ++ ;
            cur = 1;
            mxt = t[i] + k;
            CC;
        }
        if (cur==c)
        {
            cnt ++ ;
            cur = 1;
            mxt = t[i] + k;
            CC;
        }
        cur ++ ;
    }
    return cnt <= x;
}
int main()
{
    cin >> n>> c >> k ;
    lo = (n+c-1) / c;
    hi = n;
    lop(i,n) cin >> t[i];
    sort(t,t+n);
    while (lo < hi )
    {
        md = (lo + hi  ) /2 ;
        if (!ok(md)) lo = md + 1;
        else hi = md;
    }
    cout << hi << endl ;
}

/*
100 10 15 10
1000 500 500 20
https://ideone.com/goT6Vy//http:
codeforces.com/problemset/problem/466/D

, vector < pair <int,int > >, greater <pair <int, int> >
http://codeforces.com/contest/295/submission/17929453
freopen("output.txt","w",stdout);
freopen("input.txt","r",stdin);

ll pwmod(ll a, ll n, ll mod)
{
    ll ret = 1;
    while (n)
    {
        if (n & 1) ret = ret * a % mod;
        a = a * a % mod;
        n >>= 1;
    }
    return ret;
}

for (int i=0; i<n; i++)            Hi    ^_^  hack me if you can :P

int mx[]= {0,0,1,-1, 1,1,-1,-1};
int my[]= {1,-1,0,0, 1,-1,1,-1};
bool inside (int i,int j){return (i>=0 && i<n && j>=0 && j<m);}

typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> pbds;

void gcd(LL n,LL m,LL &d,LL &xx,LL &yy)
{
    if(!m)
    {
        d=n;
        xx=1;
        yy=0;
    }
    else
    {
        gcd(m,n%m,d,yy,xx);
        yy-=xx*(n/m);
    }
}

int getm(int l,int r)
{
    if(l>r)
        return N;
    if(l==r)
        return 0;
    if(l==r-1)
        return s[l] != s[r];
    if(s[l]==s[r])
        return getm(l+1,r-1);
    else
        return 1+min(getm(l+1,r),getm(l,r-1));
}

int BIT[si], a[si], n;
void update(int x, int delta)
{
    for(; x <= N; x += x&-x)
        BIT[x] += delta;
}
int query(int x)
{
    int sum = 0;
    for(; x > 0; x -= x&-x)
        sum += BIT[x];
    return sum;
}

#define sc(x)      scanf("%d",&x)
#define scl(x)     scanf("%lld",&x)
#define pr(x)      printf("%d ",x)
#define prl(x)     printf("%lld ",x)
#define prln(x)    printf("%d\n",x)
#define prlln(x)   printf("%lld\n",x)
#define scd(x)     scanf("%lf",&x)
#define prd(x)     printf("%.10f\n",x)


*/