// @EXPECTED_RESULTS@: TIME_LIMIT_EXCEEDED,WRONG_ANSWER
#include<algorithm>
#include<cassert>
#include<climits>
#include<cmath>
#include<iostream>
#include<numeric> // accumulate
#include<vector>
using namespace std;
typedef long long ll;

template<int kp>
ll min_diff(vector<ll>::iterator from, vector<ll>::iterator to, ll target) {
	if constexpr(kp == 1) {
		assert(from != to);
		vector<ll>::iterator it = upper_bound(from, to, target);
		ll res = (it == to ? LLONG_MAX : *it - target);
		if (it != from) return min(res, target - *(--it));
		else return res;
	} else {
		ll res = LLONG_MAX;
		while (distance(from, to) >= kp) {
			ll curtarget = target - *from;
			res = min(res, min_diff<kp - 1>(++from, to, curtarget));
		}
		return res;
	}
}

long double solve() {
	ll n, k, xbar;
	vector<ll>::iterator it;

	cin >> n >> k >> xbar;
	vector<ll> x(n);
	for (ll &xi : x) cin >> xi;
	for (ll &xi : x) xi -= xbar; // set xbar = 0 WLOG
	sort(x.begin(), x.end());

	ll s = accumulate(x.begin(), x.end(), 0LL);

	long double res = (long double)(s) / n; // do not remove anything
	if (k <= 0) return fabs(res);

	// O(log n)
	res = min(res, (long double)min_diff<1>(begin(x), end(x), s) / (n-1));
	if (k <= 1) return fabs(res);

	// O(n log n)
	res = min(res, (long double)min_diff<2>(begin(x), end(x), s) / (n-2));
	if (k <= 2) return fabs(res);

	// O(n^2 log n)
	res = min(res, (long double)min_diff<3>(begin(x), end(x), s) / (n-3));
	if (k <= 3) return fabs(res);

	// O(n^3 log n)
	res = min(res, (long double)min_diff<4>(begin(x), end(x), s) / (n-4));
	if (k <= 4) return fabs(res);

	assert(false); // k >= 5 not supported
	return fabs(res);
}

int main() {
	(cout << fixed).precision(6);
	cout << solve() << endl;
}