#include "bits/stdc++.h"
#define int long long
using namespace std;
using ll = long long;
using P = pair<ll, pair<ll, ll>>;
const ll INF = (1LL << 61);
ll mod = 1000000007;
struct mint {
	ll x; // typedef long long ll;
	mint(ll x = 0) :x((x%mod + mod) % mod) {}
	mint operator-() const { return mint(-x); }
	mint& operator+=(const mint a) {
		if ((x += a.x) >= mod) x -= mod;
		return *this;
	}
	mint& operator-=(const mint a) {
		if ((x += mod - a.x) >= mod) x -= mod;
		return *this;
	}
	mint& operator*=(const mint a) {
		(x *= a.x) %= mod;
		return *this;
	}
	mint operator+(const mint a) const {
		mint res(*this);
		return res += a;
	}
	mint operator-(const mint a) const {
		mint res(*this);
		return res -= a;
	}
	mint operator*(const mint a) const {
		mint res(*this);
		return res *= a;
	}
	mint pow(ll t) const {
		if (!t) return 1;
		mint a = pow(t >> 1);
		a *= a;
		if (t & 1) a *= *this;
		return a;
	}

	// for prime mod
	mint inv() const {
		return pow(mod - 2);
	}
	mint& operator/=(const mint a) {
		return (*this) *= a.inv();
	}
	mint operator/(const mint a) const {
		mint res(*this);
		return res /= a;
	}
};
istream& operator>>(istream& is, mint& a) { return is >> a.x; }
ostream& operator<<(ostream& os, const mint& a) { return os << a.x; }
long long dp[55][3000][55];
signed main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	int N, M, X, Y; cin >> N >> M >> X >> Y;
	int Z; cin >> Z;
	vector<int>A(N); for (int i = 0; i < N; i++)cin >> A[i];
	sort(A.begin(), A.end());
	int cnt = 0;
	for (int i = 0; i < N; i++) {
		if (A[i] >= X)cnt++;
	}
	bool ok = false;
	if (cnt == 0)ok = true;

	if (N - cnt < N - M) {
		
		cout << "Handicapped" << endl; return 0;
	}
	int cnt2 = 0;
	for (int i = 0; i < N; i++) {
		if (A[i] <= Y)cnt2++;
	}
	int sum = 0;
	for (int i = 0; i < N; i++) {
		if(A[i] >= X) sum += A[i];
	} 
	dp[0][0][0] = 1;
	for (int i = 0; i < N - cnt2; ++i) {
		for (int s = 0; s <= 2500; ++s) {
			for (int k = 0; k <= N; ++k) {
				if (dp[i][s][k] == 0) continue;
				if(A[cnt2 + i] < X)dp[i + 1][s][k] += dp[i][s][k];
				dp[i + 1][s + A[cnt2 + i]][k + 1] += dp[i][s][k];
			}
		}
	}
	long long res = 0;
	for (int k = 1; k <= M; ++k) {
		//if (N - cnt2 == M && k == M)continue;
		res += dp[N - cnt2][Z*k][k];
	}
	cout << res << endl;
	
	return 0;
}