Suppose we have already rolled some dice. Let $$$a_i$$$ be the number of occurrences of the number $$$i$$$, for each $$$1\leq i\leq M$$$. Let's say that the sorted list of $$$[a_1,a_2,\cdots,a_M]$$$ is the configuration of the current state.

Now we are going to do a dynamic programming approach. Let's mantain a DP table:

We want to calculate $$$DP[0,0,\cdots,0]$$$. Since $$$N\leq 50$$$, the number of possible configurations is reasonably small, so that the solution fits in time and memory limit.

Now suppose we are to compute $$$DP[C]$$$ for some configuration $$$C$$$. We should minimize the number of dice rolls, so we will try every possible strategy of the following form.

1. Predetermine a subset $$$S$$$ of the set of all possible configurations after a single roll of die.
2. Roll a die and check the current configuration. If it is in $$$S$$$, proceed to the next step. Otherwise, re-roll the die.

Our job is to determine $$$S$$$. We will sort all possible configurations $$$C'$$$ by $$$DP[C']$$$ in ascending order, while excluding those that is impossible to get to the final configuration. (Those with $$$DP[C']=\infty$$$) Try $$$S$$$ to be every prefix of this sorted list. One can easily prove that this always find the optimal strategy.

#include <bits/stdc++.h>
#define debug(...) 42

using namespace std;

int n, m, k;
map<vector<int>, double> dp;
vector<int> target;

void tc_init() {
	dp.clear();
	target.clear();
}

double solve(vector<int> &v, int depth) {
	auto it = dp.find(v);
	if (it != dp.end()) return it->second;
	double res = 1e18, cs = 0.0;
	int cnt = 0, s = 0;
	for (int i = 0; i < m; i++) {
		if (!i || v[i] != v[i - 1]) cnt = 1;
		else ++cnt;
		if ((i == m - 1 || v[i] != v[i + 1]) && v[i] < target[i]) {
			v[i]++;
			s += cnt;
			cs += solve(v, depth + 1) * cnt;
			v[i]--;
		}
	}
	return dp[v] = 1.0 * m / s + cs / s;
}

int tc_main() {
	cin >> n >> m >> k;
	target = vector<int>(m, 0);
	vector<int> v(m, 0);
	for (int i = m - k; i < m; i++) {
		cin >> target[i];
	}
	dp[target] = 0.0;
	cout << setprecision(10) << solve(v, 0) << "\n";
	return 0;
}

int main() {
	ios::sync_with_stdio(false);
	cin.tie(NULL);
	int tc;
	cin >> tc;
	for (int t = 1; t <= tc; t++) {
		cout << "Case #" << t << ": ";
		tc_init();
		tc_main();
	}
	return 0;
}