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<bits/stdc++.h>
using namespace std;

int n,m,l;
vector<int> a,b,c;
bool input(){
  a=b=c=vector<int>(20,0);
  cin>>n>>m>>l;
  if(n==0&&m==0&&l==0)return false;
  for(int i=0;i<l;i++){
    string tmp;
    cin>>tmp;
    switch(tmp[0]){
      case 'a':
        a[tmp.size()-1]++;
        break;
      case 'b':
        b[tmp.size()-1]++;
        break;
      case 'c':
        c[tmp.size()-1]++;
        break;
    }
  }
  return true;
}
typedef long long ll;
ll nCk(int n,int k){
  if(n<k){
    return 0;
  }
  ll res = 1;
  for(int i=n-k+1;i<=n;i++)res*=i;
  for(int i=1;i<=k;i++)res/=i;
  return res;
}


bool check_sub(vector<int> h,int ai,int bi,int ci){
  vector<int> need(20,0);
  for(int i=0;i<a.size();i++){
    need[h[ai]+i] += a[i];
    need[h[bi]+i] += b[i];
    need[h[ci]+i] += c[i];
  }
  vector<int> now(20,0);
  for(int i=0;i<h.size();i++){
    now[h[i]]++;
  }
  for(int i=0;i<need.size();i++){
    if(need[i]>now[i])return false;
    if(now[i]>n)return false;
  }
  return true;
}

bool check(vector<int> h){
  for(int i=0;i<h.size();i++){
    for(int j=0;j<h.size();j++){
      for(int k=0;k<h.size();k++){
        if(check_sub(h,i,j,k))return true;;
      }
    }
  }
  return false;
}

ll count(vector<int> h,int b){
  vector<int> now(20,0);
  for(int i=0;i<h.size();i++){
    now[h[i]]++;
  }
  ll res = 1;
  for(int i=0;i<now.size();i++){
    res*=nCk(n,now[i]);
  }
  const int arr = m - (h[h.size()-1]-b+1) + 1;
  res *= nCk(arr,b+1);
  return res;
}

vector<int> hands;
ll req(int n,int b){
  if(!(hands[n]<m))return 0;
  if(n==hands.size()-1){
    if(check(hands)){
      return count(hands,b);
    }else{
      return 0;
    }
  }
  ll res = 0;
  hands[n+1] = hands[n];
  res += req(n+1,b);
  hands[n+1] = hands[n]+1;
  res += req(n+1,b);
  hands[n+1] = hands[n]+2;
  res += req(n+1,b+1);
  return res;
}
ll solve(){
  hands = vector<int>(l);
  hands[0]=0;
  return req(0,0);
}


int main(){
  while(input()){
    long long ans = solve();
    cout<<fixed<<setprecision(10)<<((double)ans)/nCk(n*m,l)<<endl;
  }
}