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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

#define ll long long
#define fw(p) for(int w=0;w<(p);w++)
#define fx(p) for(int x=0;x<(p);x++)
#define fy(p) for(int y=0;y<(p);y++)
#define fz(p) for(int z=0;z<(p);z++)
#define fyg(p,g) for(int y=(g);y<(p);y++)
#define fzg(p,g) for(int z=(g);z<(p);z++)
#define ce(d) cout<<d<<endl;
#define vecp(p) int aa;cin>>aa;(p).push_back(aa);
#define vecpl(p) long long aa;cin>>aa;(p).push_back(aa);
#define vecps(p) string aa;cin>>aa;(p).push_back(aa);
#define vecp2(p) cin>>aa;(p).push_back(aa);
#define vecpl2(p) long long a	b;cin>>ab;(p).push_back(ab);
#define vecps2(p) string ab;cin>>ab;(p).push_back(ab);
#define set0(k,n) for(int nn=0;nn<(n);nn++){ (k).push_back(0); }
#define sorts(c) sort((c).begin(),(c).end());
#define reverses(c) reverse((c).begin(),(c).end());
#define vec(b) vector<int> (b);
#define vecl(b) vector<long long> (b);
#define vecs(b) vector<string> (b);
#define vecsize(b,size) vector<int> (b)((size));
#define pb(b,a) (b).push_back((a));
#define doublece(a,b) cout<<(a)<<' '<<(b)<<endl;
#define pairs(s) vector<pair<int,int>> (s);
#define pairsl(s) vector<pair<ll,ll>> (s);
#define pairss(s) vector<pair<string,string>> (s);
#define pairsp(s) int aa,bb;cin>>aa>>bb;(s).push_back(make_pair(aa,bb));
#define pairspl(s) int aa,bb;cin>>aa>>bb;(s).push_back(make_pair(aa,bb));
#define pairsps(s) int aa,bb;cin>>aa>>bb;(s).push_back(make_pair(aa,bb));
#define pairsREV(s) (s).push_back(make_pair(bb,aa));
#define pairslREV(s) (s).push_back(make_pair(bb,aa));
#define pairssREV(s) (s).push_back(make_pair(bb,aa));
#define MOD 1000000007

int main()
{
ll N,A,B;
	cin >> N>>A>>B;
	vecl(H);
	fx(N) {
		vecpl(H);
	}
	A -= B;
	ll left = 0, right = 2000000000;
	ll mid;
	fx(50) {
		mid = (left + right +1) / 2;
		int k = mid;
		fy(N) {
			if (H[y] >mid*B) {
				ll tmp = (H[y] - mid * B);
				k -= (tmp + A - 1) / A;
				//at = mid- k;
			}
		}

		if (k>=0) {
			right = mid;
		}
		else {
			left = mid;
		}
	}

			ce(right)

    return 0;
}