题干：

You are policeman and you are playing a game with Slavik. The game is turn-based and each turn consists of two phases. During the first phase you make your move and during the second phase Slavik makes his move.

There are nn doors, the ii-th door initially has durability equal to aiai.

During your move you can try to break one of the doors. If you choose door ii and its current durability is bibi then you reduce its durability to max(0,bi−x)max(0,bi−x) (the value xxis given).

During Slavik's move he tries to repair one of the doors. If he chooses door ii and its current durability is bibi then he increases its durability to bi+ybi+y (the value yyis given). Slavik cannot repair doors with current durability equal to 00.

The game lasts 1010010100 turns. If some player cannot make his move then he has to skip it.

Your goal is to maximize the number of doors with durability equal to 00 at the end of the game. You can assume that Slavik wants to minimize the number of such doors. What is the number of such doors in the end if you both play optimally?

Input

The first line of the input contains three integers nn, xx and yy (1≤n≤1001≤n≤100, 1≤x,y≤1051≤x,y≤105) — the number of doors, value xx and value yy, respectively.

The second line of the input contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤1051≤ai≤105), where aiaiis the initial durability of the ii-th door.

Output

Print one integer — the number of doors with durability equal to 00 at the end of the game, if you and Slavik both play optimally.

Examples

Input

6 3 2
2 3 1 3 4 2

Output

6

Input

5 3 3
1 2 4 2 3

Output

2

Input

5 5 6
1 2 6 10 3

Output

2

Note

Clarifications about the optimal strategy will be ignored.

题目大意：

给你一个含有N个数的数组，每一个元素代表一个门的当前防御值

每一次你可以对门攻击x个点数，而一个神仙可以对门进行y个点数的防御值提升（也可以高于初始防御值）。

当一次你对门的攻击使这个门的防御值小于等于0的时候，这个门就坏掉了，神仙也没法修复了。

问：当你和神仙都采取最优的策略的时候，你最多可以砸坏几个门？

解题报告：

1.    x>y ，这样的话，每一个门你都可以给砸坏。

2：当x<=y，这样你的最优策略就是每一次去砸那些当前防御值比你的攻击力x值小的门，一次就可以给砸坏，

而神仙的最优策略使去提升那些当前防御值比你的攻击力x值小的门，一次来减少你的数量。当然，如果你砸那些不能一次砸坏的，他就来修复，所以这并不影响答案。

AC代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define ll long long
#define pb push_back
#define pm make_pair
using namespace std;
const int MAX = 2e5 + 5;
int n,x,y,a[MAX],ans;
int main() {
	cin>>n>>x>>y;
	ans=0;
	for(int i=1; i<=n; i++) {
		scanf("%d",&a[i]);
		if(a[i]<=x) ans++;
	}
	if(x>y) printf("%d\n",n);
	if(x<=y) {
		if(ans%2==0) printf("%d\n",ans/2);
		else printf("%d\n",ans/2+1);
	}
	return 0;
}