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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
#define REP(i, n) for(int i = 0;i < n;i++)
#define REPR(i, n) for(int i = n;i >= 0;i--)
#define FOR(i, m, n) for(int i = m;i < n;i++)
#define itrfor(itr,A) for(auto itr = A.begin(); itr !=A.end();itr++)
typedef int64_t llong;
char moji[26]={'a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z'};
char moji2[26]={'A','B','C','D','E','F','G','H','I','J','K','L','M','N','O','P','Q','R','S','T','U','V','W','X','Y','Z'};
char moji3[10]={'0','1','2','3','4','5','6','7','8','9'};
#define Sort(a) sort(a.begin(),a.end());
#define Reverse(a) reverse(a.begin(),a.end());
#define ld PRId64
#define n_MAX int(1e5)





int main(){
    int n;
    cin >> n;
    llong A,B;
    cin >> A >> B;
    
    llong tai[n_MAX + 5];
    REP(i,n) scanf("%ld",&tai[i]);
    llong left = -1;
    llong right = 1e9;
    llong mid,kijun,cnt;
    while(left + 1 < right){
        mid=(left + right) /2;
        kijun= B * mid;
        cnt=0;
        REP(i,n){
            cnt += max((tai[i] - kijun)/(A-B),llong(0));
            if( (tai[i] - kijun)%(A-B)!=0 and tai[i] - kijun >0) cnt +=1;
        }
        if(cnt > mid) left = mid;
        else right = mid;
    }
    cout << right << endl;

}