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 C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
#include<math.h>
 
void QSort(long long x[ ], long long left, long long right);
void Swap(long long x[ ], long long i, long long j);
 
 
 
int main(){
  
  long long n,a,b;

  scanf("%lld%lld%lld",&n,&a,&b);
  
  long long h[n];
  
  for(int i=0;i<n;i++){
    scanf("%lld",&h[i]);
  }

  QSort(h,0,n-1);

  
  long long ans=0;

  long long kosu=n;

  while(kosu>0){
    h[0]=h[0]-a;
    if(h[0]<=0){
      kosu=kosu-1;
    }
    for(long long i=1;i<n;i++){
      h[i]=h[i]-b;
      if(h[i]<=0){
	kosu--;
      }
    }
    QSort(h,0,n-1);
    ans++;
  }
      
    
    
  printf("%lld\n",ans);
  
  
  return 0;
}
 
  /* クイックソートを行う */
void QSort(long long x[ ], long long left, long long right)
{
    long long i, j;
    long long pivot;
 
    i = left;                      /* ソートする配列の一番小さい要素の添字 */
    j = right;                     /* ソートする配列の一番大きい要素の添字 */
 
    pivot = x[(left + right) / 2]; /* 基準値を配列の中央付近にとる */
 
    while (1) {                    /* 無限ループ */
 
        while (x[i] > pivot)       /* pivot より大きい値が */
            i++;                   /* 出るまで i を増加させる */
 
        while (pivot > x[j])       /* pivot より小さい値が */
            j--;                   /*  出るまで j を減少させる */
        if (i >= j)                /* i >= j なら */
            break;                 /* 無限ループから抜ける */
 
        Swap(x, i, j);             /* x[i] と x[j]を交換 */
        i++;                       /* 次のデータ */
        j--;
    }
   
 
    if (left < i - 1)              /* 基準値の左に 2 以上要素があれば */
        QSort(x, left, i - 1);     /* 左の配列を Q ソートする */
    if (j + 1 <  right)            /* 基準値の右に 2 以上要素があれば */
        QSort(x, j + 1, right);    /* 右の配列を Q ソートする */
}
 
  /* 配列の要素を交換する */
void Swap(long long x[ ], long long i, long long j)
{
    long long temp;
 
    temp = x[i];
    x[i] = x[j];
    x[j] = temp;
}