Problem:
Score : 700 points

Problem Statement

N problems are proposed for an upcoming contest. Problem i has an initial integer score of A_i points.

M judges are about to vote for problems they like. Each judge will choose exactly V problems, independently from the other judges,
and increase the score of each chosen problem by 1.

After all M judges cast their vote, the problems will be sorted in non-increasing order of score, and the first P problems will be chosen for the problemset.
Problems with the same score can be ordered arbitrarily, this order is decided by the chief judge.

How many problems out of the given N have a chance to be chosen for the problemset?

Constraints

2 \le N \le 10^5

1 \le M \le 10^9

1 \le V \le N - 1

1 \le P \le N - 1

0 \le A_i \le 10^9

Input

Input is given from Standard Input in the following format:

N M V P
A_1 A_2 ... A_N

Output

Print the number of problems that have a chance to be chosen for the problemset.

Sample Input 1

6 1 2 2
2 1 1 3 0 2

Sample Output 1

5

If the only judge votes for problems 2 and 5, the scores will be 2 2 1 3 1 2.
The problemset will consist of problem 4 and one of problems 1, 2, or 6.

If the only judge votes for problems 3 and 4, the scores will be 2 1 2 4 0 2.
The problemset will consist of problem 4 and one of problems 1, 3, or 6.

Thus, problems 1, 2, 3, 4, and 6 have a chance to be chosen for the problemset. On the contrary, there is no way for problem 5 to be chosen.

Sample Input 2

6 1 5 2
2 1 1 3 0 2

Sample Output 2

3

Only problems 1, 4, and 6 have a chance to be chosen.

Sample Input 3

10 4 8 5
7 2 3 6 1 6 5 4 6 5

Sample Output 3

8

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

#include <bits/stdc++.h>


using namespace std;

typedef unsigned long long ull;
typedef long long ll;
typedef long double ld;


#define int ll
#define rand shittttty_shit
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
mt19937_64 rng_64(chrono::steady_clock::now().time_since_epoch().count());


typedef pair<int, int> pii;
typedef pair<string, string> pst;
typedef pair<pii, pii> piii;
typedef vector<piii> vpii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;
typedef vector< vi > vvi;
typedef vector< vvi > vvvi;
typedef vector<pii> vpi;
typedef vector<short> vs;
typedef vector<vs> vvs;
typedef vector<vvs> vvvs;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef vector<vvl> vvvl;
typedef vector<ld> vld;
typedef vector<vld> vvld;
typedef vector<vvld> vvvld;
typedef vector<string> vst;
typedef vector<vst> vvst;
typedef pair<ld, ld> pld;
typedef pair<pld, pld> pldd;
typedef vector<pld> vpd;
typedef complex<double> base;


#define inmin(a, b) a = min(a, (b))
#define inmax(a, b) a = max(a, (b))
#define my_abs(a) (((a) >= 0) ? (a) : -(a))
#define mp(a, b) make_pair(a, (b))
#define ALL(a) a.begin(),a.end()
#define RALL(a) a.rbegin(),a.rend()
#define sqr(x) ((x) * (x))
#define fori(i, n) for(int i = 0; i < int(n); ++i)
#define SZ(a) ((int)((a).size()))
#define MODIK(a) (a >= M ? a - M : a)
#define triple(T) tuple<T, T, T>
#define quad(T) tuple<T, T, T, T>
#define watch(x) cout << (#x) << " = " << x << endl;
#ifdef MAX_HOME
#define cerr cout
#else
#define cerr if (false) cerr
#endif


const double PI = 2 * acos(0.0);

const string DIGITS = "0123456789";
const string ALPH = "abcdefghijklmnopqrstuvwxyz";

istream &operator >> (istream &in, pii &a)
{
    in >> a.first >> a.second;
    return in;
}


ostream &operator << (ostream &out, pii &a)
{
    out << a.first << ' ' << a.second;
    return out;
}


istream &operator >> (istream &in, pld &a)
{
    in >> a.first >> a.second;
    return in;
}


ostream &operator << (ostream &out, pld &a)
{
    out << a.first << ' ' << a.second;
    return out;
}


template <class T0, class T1>
inline ostream & operator << (ostream &out, pair<T0, T1> &a) {
    return out << "{" << a.first << ", " << a.second << "}";
}

template <class T0, class T1, class T2>
inline ostream & operator << (ostream &out, tuple<T0, T1, T2> &a) {
    return out << "{" << get<0>(a) << ", " << get<1>(a) << ", " << get<2>(a) << "}";
}

template <class T0, class T1, class T2, class T3>
inline ostream & operator << (ostream &out, tuple<T0, T1, T2, T3> &a) {
    return out << "{" << get<0>(a) << ", " << get<1>(a) <<
               ", " << get<2>(a) << ", " <<  get<3>(a) << "}";
}

template<class T>
inline ostream & operator << (ostream &out, vector<T> &a) {
    out << "[";
    fori (i, a.size())
        out << a[i] << vector<string>{", ", "]  "}[i + 1 == a.size()];
    return out;
}


void print(vi kek, int k)
{
    ofstream cout("output.txt");
    cout << SZ(kek) << ' ' << k << '\n';
    for (auto p : kek) {
        cout << p << ' ';
    }
    cout.close();
    exit(0);
}


inline bool break_time(ld time)
{
    return clock() > CLOCKS_PER_SEC * time;
}


ld get_rand_ld(ld l, ld r)
{
    uniform_real_distribution<double> gen(l, r);
    return gen(rng);
}


void smain();

signed main() {
    ios::sync_with_stdio(0);
    cin.tie(0); cout.tie(0);

#ifdef ART_HOME
    freopen("input.txt", "r", stdin);
    clock_t start = clock();
#endif
    cout << setprecision(4) << fixed;
    smain();
#ifdef ART_HOME
    cout << "\n\n\nTOTAL EXECUTION TIME: " << float( clock () - start ) /  CLOCKS_PER_SEC << endl;
#endif
}


const int M1 = 1000000093, p1 = 27162;
const int M2 = 1000000087, p2 = 241;


const int N = 50, M = 998244353, oo = 2e18 + 100;
const ld PEPS = 1e-9;


void smain() {
    int n, m, v, p;
    cin >> n >> m >> v >> p;

    vi kek(n);
    for (int i = 0; i < n; ++i)
        cin >> kek[i];

    stable_sort(ALL(kek));
    vi flex = kek;
    for (int i = 1; i < n; ++i)
        kek[i] += kek[i - 1];

    int ans = p;

    for (int i = 0; i + p < n; ++i)
    {
        int cnt = m * max(0LL, v - p - i);

        if (flex[i] + m < flex[n - p])
            continue;

        if (!cnt)
        {
            ans++;
            continue;
        }

        int len = (n - p - i);
        int c = m * len + flex[i] * len - (kek[n - p] - kek[i]);
        if (c >= cnt)
            ans++;
    }

    cout << ans;
}