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;

#define ALL(x) (x).begin(), (x).end()
#define AND(x,y,z) 
#define DUMP(xs) for (auto x:xs) cout<<x<<' ';cout<<endl
#define DUMPP(xs) for (auto x:xs) cout<<'('<<x.first<<','<<x.second<<')';cout<<endl
#define FOR(i,a,b) for (ll i=(ll)(a);i<(ll)(b);++i)
#define OUT(x) cout<<x<<endl
#define REP(i,n) FOR(i,0,n)
#define YES(cond) cond ? OUT("YES") : OUT("NO")
#define Yes(cond) cond ? OUT("Yes") : OUT("No")

template<class T> T gcd(const T x,const T y){if(!y){return x;}return gcd(y,x%y);}
template<class T> T lcm(const T x,const T y){return x/gcd(x,y)*y;};
template<class T> T modpow(const T x,const T n,const T mod) {
  T res=1;while(n>0){if(n%2)res=res*x%mod;x=x*x%mod;n/=2;};return res;
}
template<class T> bool alltrue(const T &a){return all_of(ALL(a),[](bool x){return x;});}
template<class T> bool anytrue(const T &a){return any_of(ALL(a),[](bool x){return x;});}
template<class T> bool contain(const T &a,const T &b){for(auto x:b){if (a.find(x)==a.end()){return false;}}return true;}
template<class T> bool mmax(T &m,const T q){ if (m < q) {m = q; return true;} else return false; }
template<class T> bool mmin(T &m,const T q){ if (m > q) {m = q; return true;} else return false; }
template<class T> int siz(const T &a) {return (int)a.size();}
template<class T> set<T> devisers(T n) {
  set<T> res;for(T i=1;i<(T)sqrt(n)+1;i++){if(n%i==0){res.insert(i);res.insert(n/i);}}return res;
}
template<class T> void concat(T &a, const T &b){a.insert(a.end(),b.begin(),b.end());}
template<class T> void getand(const T &a,const T &b,T &c) {
  set_intersection(a.begin(),a.end(),b.begin(),b.end(),inserter(c,c.end()));
}
template<class T> void getor(const T &a,const T &b,T &c) {
  set_union(a.begin(),a.end(),b.begin(),b.end(),inserter(c,c.end()));
}

typedef long long ll;
typedef pair<ll, ll> P;

const ll MOD=1e9+7;
const int MAX=1e5+100;

int main() {
  // 採用されうる問題数
  // ASを降順にした時、上位P-1番目までは確定
  // P番目以降にについて、スコアごとに二分探索していく
  ll N,M,V,P;cin>>N>>M>>V>>P;
  vector<ll> AS(N);
  REP(i,N) cin>>AS[i];

  // ASの降順取得
  sort(ALL(AS));
  reverse(ALL(AS));
  // DUMP(AS);

  vector<ll> dupas(AS);
  sort(ALL(dupas));
  auto checker = [&](ll x) {
    // Mを足してもP番目の数に到達できない場合
    if (x+M<AS[P-1]) return false;

    // x以降の数6|55...0
    ll smaller_cnt = upper_bound(ALL(dupas),x)-dupas.begin();
    ll tmp = 0;
    FOR(i,P-1,N-smaller_cnt) tmp += x+M-AS[i];
    ll baffer = ((P-1)+smaller_cnt)*M+tmp;
    return baffer>=M*V;
  };

  // スコアが最小でもOKの場合
  if (checker(AS.back())) {
    OUT(N);
    return 0;
  }

  ll left=AS.back(),right=AS[P-1];
  while (left<right) {
    ll mid = (left+right)/2;
    checker(mid) ? right=mid : left=mid+1;
  }
  // OUT(left);

  ll imp_cnt = lower_bound(ALL(dupas),left)-dupas.begin();
  OUT(N-imp_cnt);

  return 0;
}