Problem: You have a deck of N × M cards. Each card in the deck has a rank. The range of ranks is 1 through M, and the deck includes N cards of each rank.

We denote a card with rank m by m here.

You can draw a hand of L cards at random from the deck. If the hand matches the given pattern, some bonus will be rewarded. A pattern is described as follows.


hand_pattern = card_pattern1 ' ' card_pattern2 ' ' ... ' ' card_patternL
card_pattern = '*' | var_plus
var_plus = variable | var_plus '+'
variable = 'a' | 'b' | 'c'


hand_pattern      A hand matches the hand_pattern if each card_pattern in the hand_pattern matches with a distinct card in the hand.  card_pattern      If the card_pattern is an asterisk ('*'), it matches any card. Characters 'a', 'b', and 'c' denote variables and all the occurrences of the same variable match cards of the same rank. A card_pattern with a variable followed by plus ('+') characters matches a card whose rank is the sum of the rank corresponding to the variable and the number of plus characters. You can assume that, when a hand_pattern includes a card_pattern with a variable followed by some number of plus characters, it also includes card_patterns with that variable and all smaller numbers (including zero) of plus characters. For example, if 'a+++' appears in a hand_pattern, card_patterns 'a', 'a+', and 'a++' also appear in the hand_pattern.

There is no restriction on which ranks different variables mean. For example, 'a' and 'b' may or may not match cards of the same rank.

We show some example hand_patterns. The pattern


a * b a b


matches the hand:


3 3 10 10 9


with 'a's and 'b's meaning 3 and 10 (or 10 and 3), respectively. This pattern also matches the following hand.


3 3 3 3 9


In this case, both 'a's and 'b's mean 3. The pattern


a a+ a++ a+++ a++++


matches the following hand.


4 5 6 7 8


In this case, 'a' should mean 4.

Your mission is to write a program that computes the probability that a hand randomly drawn from the deck matches the given hand_pattern.

Input

The input is a sequence of datasets. Each dataset is formatted as follows.

> N M L
>  card_pattern1 card_pattern2 ... card_patternL

The first line consists of three positive integers N, M, and L. N indicates the number of cards in each rank, M indicates the number of ranks, and L indicates the number of cards in a hand. N, M, and L are constrained as follows.

> 1 ≤ N ≤ 7
>  1 ≤ M ≤ 60
>  1 ≤ L ≤ 7
>  L ≤ N × M
>

The second line describes a hand_pattern.

The end of the input is indicated by a line containing three zeros separated by a single space.

Output

For each dataset, output a line containing a decimal fraction which means the probability of a hand matching the hand_pattern.

The output should not contain an error greater than 10−8.

No other characters should be contained in the output.

Sample Input


1 1 1
a
3 3 4
a+ * a *
2 2 3
a a b
2 2 3
* * *
2 2 3
* b b
2 2 2
a a
2 3 3
a a+ a++
2 6 6
a a+ a++ b b+ b++
4 13 5
a a * * *
4 13 5
a a b b *
4 13 5
a a a * *
4 13 5
a a+ a++ a+++ a++++
4 13 5
* * * * *
4 13 5
a a a b b
4 13 5
a a a a *
7 60 7
a b a b c c *
7 60 7
* * * * * * *
7 60 7
a a+ a++ a+++ a++++ a+++++ a++++++
1 14 4
b a+ a a
0 0 0


Output for the Sample Input


1.0000000000
0.8809523810
1.0000000000
1.0000000000
1.0000000000
0.3333333333
0.4000000000
0.1212121212
0.4929171669
0.0492196879
0.0228091236
0.0035460338
1.0000000000
0.0014405762
0.0002400960
0.0002967709
1.0000000000
0.0000001022
0.0000000000






Example

Input

1 1 1
a
3 3 4
a+ * a *
2 2 3
a a b
2 2 3
* * *
2 2 3
* b b
2 2 2
a a
2 3 3
a a+ a++
2 6 6
a a+ a++ b b+ b++
4 13 5
a a * * *
4 13 5
a a b b *
4 13 5
a a a * *
4 13 5
a a+ a++ a+++ a++++
4 13 5
* * * * *
4 13 5
a a a b b
4 13 5
a a a a *
7 60 7
a b a b c c *
7 60 7
* * * * * * *
7 60 7
a a+ a++ a+++ a++++ a+++++ a++++++
1 14 4
b a+ a a
0 0 0


Output

1.0000000000
0.8809523810
1.0000000000
1.0000000000
1.0000000000
0.3333333333
0.4000000000
0.1212121212
0.4929171669
0.0492196879
0.0228091236
0.0035460338
1.0000000000
0.0014405762
0.0002400960
0.0002967709
1.0000000000
0.0000001022
0.0000000000

Solution (in CPP):

Solution:
#include <iostream>
#include <string>
#include <cstdio>
#include <algorithm>
#include <vector>
#include <utility>
using namespace std;

typedef long long LL;

LL cmb[440][10];
int m, n, L;
vector<vector<int> > pts;
vector<int> hd;

bool empty(const vector<int> &v){
	return v.empty();
}

void check(int k, const vector<int> &v){
	if(k == pts.size()){
		throw 0;
	}
	vector<int> w;
	int sz = (int)v.size() - (int)pts[k].size();
	for(int i = 1; i <= sz; ++i){
		bool ok = true;
		for(size_t j = 0; j < pts[k].size(); ++j){
			if(v[i + j] < pts[k][j]){
				ok = false;
			}
		}
		if(ok){
			w = v;
			for(size_t j = 0; j < pts[k].size(); ++j){
				w[i + j] -= pts[k][j];
			}
			check(k + 1, w);
		}
	}
}

LL dfs1(int rest){
	LL res = 0;
	if(hd.size() > m + 1){ return 0; }
	if(rest == 0){
		try{
			check(0, hd);
		}
		catch(int){
			res = 1;
			int zero = 0;
			for(size_t i = 1; i < hd.size(); ++i){
				if(hd[i] == 0){
					++zero;
				}
				else{
					res *= cmb[n][hd[i]];
				}
			}
			if(zero){
				int t = hd.size() - zero - 1;
				res *= cmb[m - t - 1][zero - 1];
			}
			else if(hd.size() != m + 1){
				res = 0;
			}
		}
	}
	if(rest > 0 || hd.back()){
		for(int i = !hd.back(); i <= rest && i <= n; ++i){
			hd.push_back(i);
			res += dfs1(rest - i);
			hd.pop_back();
		}
	}
	return res;
}

int main(){
	cmb[0][0] = 1;
	for(int i = 1; i < 440; ++i){
		cmb[i][0] = 1;
		for(int j = 1; j < 10; ++j){
			cmb[i][j] = cmb[i - 1][j] + cmb[i - 1][j - 1];
		}
	}
	
	string s;
	while(cin >> n >> m >> L, n){
		pts.assign(3, vector<int>());
		for(int i = 0; i < L; ++i){
			cin >> s;
			if(s[0] != '*'){
				vector<int> &v = pts[s[0] - 'a'];
				if(v.size() < s.size()){
					v.resize(s.size());
				}
				++v[s.size() - 1];
			}
		}

		pts.erase(remove_if(pts.begin(), pts.end(), empty), pts.end());
		bool ok = true;
		for(size_t i = 0; i < pts.size(); ++i){
			if(pts[i].size() > m){
				ok = false;
			}
		}
		LL cnt = 0;
		if(ok){
			hd.assign(1, -1);
			cnt = dfs1(L);
		}
		double ans = cnt / double(cmb[n * m][L]);
		printf("%.15f\n", ans);
	}
}