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 n doors, the i-th door initially has durability equal to a_i.

During your move you can try to break one of the doors. If you choose door i and its current durability is b_i then you reduce its durability to max(0, b_i - x) (the value x is given).

During Slavik's move he tries to repair one of the doors. If he chooses door i and its current durability is b_i then he increases its durability to b_i + y (the value y is given). Slavik cannot repair doors with current durability equal to 0.

The game lasts 10^{100} 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 0 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 n, x and y (1 ≤ n ≤ 100, 1 ≤ x, y ≤ 10^5) — the number of doors, value x and value y, respectively.

The second line of the input contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^5), where a_i is the initial durability of the i-th door.

Output

Print one integer — the number of doors with durability equal to 0 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.
#include<bits/stdc++.h> /*
#include <ext/pb_ds/assoc_container.hpp> // Common file
#include <ext/pb_ds/tree_policy.hpp> // Including */
using namespace std;
//using namespace __gnu_pbds;

//typedefs

typedef long long ll;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<vi> vvi;
typedef vector<vl> vvl;
typedef pair<int,int> pii;
typedef pair<double, double> pdd;
typedef pair<ll, ll> pll;
typedef vector<pii> vii;
typedef vector<pll> vll;
typedef vector<int>::iterator vit;
typedef set<int>::iterator sit;

/*
template<typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
template<typename F, typename S>
using ordered_map = tree<F, S, less<F>, rb_tree_tag, tree_order_statistics_node_update>;
*/

//#Defines
#define rep(i,a,b)	for(i=a;i<=b;i++)
#define repR(i,a,b)	for(i=a;i>=b;i--)
//#define pb push_back
#define pb emplace_back
#define F first
#define S second
#define mp make_pair
#define all(c)		c.begin(),c.end()
#define endl '\n'
#define pf printf
#define sf scanf
//#define left __left
//#define right __right
//#define tree __tree
#define MOD 1000000007
//#define harmonic(n) 0.57721566490153286l+log(n)

#define RESET(a,b)	memset(a,b,sizeof(a))
#define gcd(a,b) __gcd(a,b)
#define lcm(a,b) (a*(b/gcd(a,b)))
#define sqr(a) ((a) * (a))

#define optimize() ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define fraction() cout.unsetf(ios::floatfield); cout.precision(10); cout.setf(ios::fixed,ios::floatfield);

const double PI = acos(-1);
const double eps = 1e-9;
const int inf = 2000000000;
const ll infLL = 9000000000000000000;


//Bit Operations
inline bool checkBit(ll n, int i) { return n&(1LL<<i); }
inline ll setBit(ll n, int i) { return n|(1LL<<i);; }
inline ll resetBit(ll n, int i) { return n&(~(1LL<<i)); }

int fx[] = {0, 0, +1, -1};
int fy[] = {+1, -1, 0, 0};
//int dx[] = {+1, 0, -1, 0, +1, +1, -1, -1};
//int dy[] = {0, +1, 0, -1, +1, -1, +1, -1};


//Inline functions

inline bool EQ(double a, double b) { return fabs(a-b) < 1e-9; }
inline bool isLeapYear(ll year) { return (year%400==0) || (year%4==0 && year%100!=0); }
inline void normal(ll &a) { a %= MOD; (a < 0) && (a += MOD); }
inline ll modMul(ll a, ll b) { a %= MOD, b %= MOD; normal(a), normal(b); return (a*b)%MOD; }
inline ll modAdd(ll a, ll b) { a %= MOD, b %= MOD; normal(a), normal(b); return (a+b)%MOD; }
inline ll modSub(ll a, ll b) { a %= MOD, b %= MOD; normal(a), normal(b); a -= b; normal(a); return a; }
inline ll modPow(ll b, ll p) { ll r = 1; while(p) { if(p&1) r = modMul(r, b); b = modMul(b, b); p >>= 1; } return r; }
inline ll modInverse(ll a) { return modPow(a, MOD-2); }
inline ll modDiv(ll a, ll b) { return modMul(a, modInverse(b)); }
inline bool isInside(pii p,ll n,ll m)	{ return (p.first>=0&&p.first<n&&p.second>=0&&p.second<m); }
inline bool isInside(pii p,ll n)		{ return (p.first>=0&&p.first<n&&p.second>=0&&p.second<n); }
inline bool isSquare(ll x) 				{ ll s = sqrt(x);	return (s*s==x); }
inline bool isFib(ll x) 				{ return isSquare(5*x*x+4)|| isSquare(5*x*x-4); }
inline bool isPowerOfTwo(ll x)			{ return ((1LL<<(ll)log2(x))==x); }


struct func
{
	//this is a sample overloading function for sorting stl
	bool operator()(pii const &a, pii const &b)	
	{
		if(a.F==b.F)
			return (a.S<b.S);
		return (a.F<b.F);
	}
};

//Prime Number Generator

/*
#define M 100000000
int marked[M/64 + 2];

#define on(x) (marked[x/64] & (1<<((x%64)/2)))
#define mark(x)  marked[x/64] |= (1<<((x%64)/2))

vl prime;

bool isPrime(int num) {
  return num > 1 && (num == 2 || ((num & 1) && !on(num))); 
}

void sieve(ll n) {
	for (ll i = 3; i * i < n; i += 2) {
		if (!on(i)) {
			for (ll j = i * i; j <= n; j += i + i) {
				mark(j);
			}
		}
	}
	prime.pb(2);
	for(ll i = 3; i <= n; i += 2) {
		if(!on(i))	prime.pb(i);	
	}
}

*/

//
//debug
#ifdef tajir
template < typename F, typename S >
ostream& operator << ( ostream& os, const pair< F, S > & p ) {
	return os << "(" << p.first << ", " << p.second << ")";
}

template < typename T >
ostream &operator << ( ostream & os, const vector< T > &v ) {
	os << "{";
	for(auto it = v.begin(); it != v.end(); ++it) {
		if( it != v.begin() ) os << ", ";
			os << *it;
	}
	return os << "}";
}

template < typename T >
ostream &operator << ( ostream & os, const set< T > &v ) {
	os << "[";
	for(auto it = v.begin(); it != v.end(); ++it) {
		if( it != v.begin() ) os << ", ";
			os << *it;
	}
	return os << "]";
}

template < typename T >
ostream &operator << ( ostream & os, const multiset< T > &v ) {
	os << "[";
	for(auto it = v.begin(); it != v.end(); ++it) {
		if( it != v.begin() ) os << ", ";
		os << *it;
	}
	return os << "]";
}

template < typename F, typename S >
ostream &operator << ( ostream & os, const map< F, S > &v ) {
	os << "[";
	for(auto it = v.begin(); it != v.end(); ++it) {
		if( it != v.begin() ) os << ", ";
		os << it -> first << " = " << it -> second ;
		}
	return os << "]";
}

#define dbg(args...) do {cerr << #args << " : "; faltu(args); } while(0)

clock_t tStart = clock();
#define timeStamp dbg("Execution Time: ", (double)(clock() - tStart)/CLOCKS_PER_SEC)

void faltu () {
	cerr << endl;
}

template <typename T>
void faltu( T a[], int n ) {
	for(int i = 0; i < n; ++i) cerr << a[i] << ' ';
	cerr << endl;
}

template <typename T, typename ... hello>
void faltu( T arg, const hello &... rest) {
	cerr << arg << ' ';
	faltu(rest...);
}
#else
#define dbg(args...)
#endif // tajir

int main()
{
	#ifdef tajir
		freopen("input.txt", "r", stdin);
	#else
		// online submission
	#endif
	optimize();
	
	int n,x,y;

	int a[10005];

	cin >> n >> x >> y;

	for(int i = 0; i < n; ++i)
		cin >> a[i];

	if(x > y)
		cout << n << endl;
	else {
		int l=0;

		for(int i = 0; i < n; ++i)
			l += (a[i]<=x);
		int ans = ceil((double)l / 2.0);

		cout << ans << endl;
	}
	
	return 0;
}






//?
