Problem:
Score : 400 points

Problem Statement

You are going out for a walk, when you suddenly encounter N monsters. Each monster has a parameter called health, and the health of the i-th monster is h_i at the moment of encounter. A monster will vanish immediately when its health drops to 0 or below.

Fortunately, you are a skilled magician, capable of causing explosions that damage monsters. In one explosion, you can damage monsters as follows:

Select an alive monster, and cause an explosion centered at that monster. The health of the monster at the center of the explosion will decrease by A, and the health of each of the other monsters will decrease by B. Here, A and B are predetermined parameters, and A > B holds.

At least how many explosions do you need to cause in order to vanish all the monsters?

Constraints

All input values are integers.

1 ≤ N ≤ 10^5

1 ≤ B < A ≤ 10^9

1 ≤ h_i ≤ 10^9

Input

Input is given from Standard Input in the following format:

N A B
h_1
h_2
:
h_N

Output

Print the minimum number of explosions that needs to be caused in order to vanish all the monsters.

Sample Input 1

4 5 3
8
7
4
2

Sample Output 1

2

You can vanish all the monsters in two explosion, as follows:

First, cause an explosion centered at the monster with 8 health. The healths of the four monsters become 3, 4, 1 and -1, respectively, and the last monster vanishes.

Second, cause an explosion centered at the monster with 4 health remaining. The healths of the three remaining monsters become 0, -1 and -2, respectively, and all the monsters are now vanished.

Sample Input 2

2 10 4
20
20

Sample Output 2

4

You need to cause two explosions centered at each monster, for a total of four.

Sample Input 3

5 2 1
900000000
900000000
1000000000
1000000000
1000000000

Sample Output 3

800000000

Solution in Haskell (GHC 7.10.3) (judge verdict: Wrong Answer)

enough::Int -> Int -> [Int] -> Int -> Bool
enough a b xs n= let strong = ((filter (\x->x>0)).(map (\x -> x-(b*n)))) xs in
                 let need = (sum.(map (\x-> (div x (a-b))+if (mod x (a-b))> 0 then 1 else 0))) strong in
                  need<=n
--a以上　b　未満で探索、bがありえない事が保証され、aが成り立つ事が保証されるとする。
binary_Int_dec::(Int,Int)->(Int -> Bool)->Maybe Int
binary_Int_dec (a,b) f = if a+1 == b then if f a then Just a else Nothing
                                 else let mid = div (a+b) 2 in
                                      if  f mid then binary_Int_dec (mid,b) f
                                                else binary_Int_dec (a,mid) f


--aより大、b以下で探索、bがありえる事が保証され、aがありえない事が保証されるとする。
binary_Int_inc::(Int,Int)->(Int -> Bool)->Maybe Int
binary_Int_inc (a,b) f = if a+1 == b then if f b then Just b else Nothing
                                 else let mid = div (a+b) 2 in
                                      if  f mid then binary_Int_inc (a,mid) f
                                                else binary_Int_inc (mid,b) f

solver::Int -> Int -> [Int] -> Maybe Int
solver a b xs = binary_Int_inc (1,1000000000) (enough a b xs)

n_get::Int -> IO [Int]
n_get 0 =return []
n_get n = do
  dc<-getLine
  rest<-n_get (n-1)
  return ((read dc):rest)

main::IO()
main=do
  d<-getLine
  let n:a:b:[]=map read (words d)
  dat<-n_get n
  case solver a b dat of
    Just x -> print x