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 LuaJIT (2.0.4) (judge verdict: Accepted)

local mfl = math.floor
local n, m = io.read("*n", "*n")
local v, p = io.read("*n", "*n")
local a = {}
for i = 1, n do
  a[i] = io.read("*n")
end
table.sort(a)

local function solve(x)
  local currank = n + 1 - x
  local has = m * (n - v)
  if m < a[n + 1 - p] - a[x] then
    return false
  end
  local need = 0
  local base = a[x]
  for i = x + 1, n + 1 - p do
    need = need + a[i] - base
  end
  -- print(x, need)
  return need <= has
end

if solve(1) then
  print(n)
else
  local min, max = 1, n
  while 1 < max - min do
    local mid = mfl((min + max) / 2)
    if solve(mid) then
      max = mid
    else
      min = mid
    end
  end
  print(n + 1 - max)
end