YYF is a couragous scout. Now he is on a dangerous mission which is to penetrate into the enemy's base. After overcoming a series difficulties, YYF is now at the start of enemy's famous "mine road". This is a very long road, on which there are numbers of mines. At first, YYF is at step one. For each step after that, YYF will walk one step with a probability of p, or jump two step with a probality of 1- p. Here is the task, given the place of each mine, please calculate the probality that YYF can go through the "mine road" safely.

Input

The input contains many test cases ended with EOF. 

Each test case contains two lines. 

The First line of each test case is N (1 ≤ N ≤ 10) and p (0.25 ≤ p ≤ 0.75) seperated by a single blank, standing for the number of mines and the probability to walk one step. 

The Second line of each test case is N integer standing for the place of N mines. Each integer is in the range of [1, 100000000].

Output

For each test case, output the probabilty in a single line with the precision to 7 digits after the decimal point.

Sample Input

1 0.522 0.52 4

Sample Output

0.50000000.2500000 题意:一条长路有 N (1 ≤ N ≤ 10)颗地雷,一个人走一步的概率是 p ，走两步的概率是 (1-p) ，然后给出 N 颗地雷的位置 ，问这个人安全走过所有地雷的概率是多少 题解:对于一个位置x，设能走到的概率是 P(x) ，那么 P(x) = P(x-1)*p + P(x-2)*(1-p) 这个数x可能很大，所以需要矩阵快速幂 然后将整个的路看成由地雷分割的 N 段路 [0 -- x1] [x1+1 -- x2] [x2+1 -- x3] ... ... 所以，他能安全过去的概率就是 N 段都能过去的连乘

1 #include 
      
        2 #include 
       
         3 #include 
        
          4 using namespace std; 5 #define MAXN 12 6  7 int n; 8 double p; 9 int bomb[MAXN];10 11 double base[2][2];12 double res[2][2];13 14 //[ p(x)   ]  =  [ p , 1-p ]^(x-1)  * [ 1 ]15 //[ p(x-1) ]     [ 1 , 0   ]          [ 0 ]16 void quick_mi(int x)17 {18     double tp[2][2];19     while (x)20     {21         if (x%2==1)22         {23             for (int i=0;i<2;i++)24                  for (int j=0;j<2;j++)25                 {26                     tp[i][j]=0;27                     for (int k=0;k<2;k++)28                         tp[i][j]+=res[i][k]*base[k][j];29                 }30             for (int i=0;i<2;i++)31                 for (int j=0;j<2;j++)32                 res[i][j]=tp[i][j];33         }34         for (int i=0;i<2;i++)35             for (int j=0;j<2;j++)36             {37                 tp[i][j]=0;38                 for (int k=0;k<2;k++)39                     tp[i][j]+=base[i][k]*base[k][j];40             }41         for (int i=0;i<2;i++)42             for (int j=0;j<2;j++)43                 base[i][j]=tp[i][j];44         x/=2;45     }46 }47 48 double Mi(int x)//处于位置1踩到位置 x 的概率49 {50     if (x==0) return 0;51     base[0][0]=p,base[0][1]=1.0-p;52     base[1][0]=1,base[1][1]=0;53     res[0][0]=1;res[0][1]=0;54     res[1][0]=0;res[1][1]=1;55     quick_mi(x-1);56     return res[0][0];57 }58 59 int main()60 {61     while (scanf("%d%lf",&n,&p)!=EOF)62     {63         for (int i=0;i