Problem:
Score : 300 points

Problem Statement

Tak has N cards. On the i-th (1 \leq i \leq N) card is written an integer x_i.
He is selecting one or more cards from these N cards, so that the average of the integers written on the selected cards is exactly A.
In how many ways can he make his selection?

Constraints

1 \leq N \leq 50

1 \leq A \leq 50

1 \leq x_i \leq 50

N,\,A,\,x_i are integers.

Partial Score

200 points will be awarded for passing the test set satisfying 1 \leq N \leq 16.

Input

The input is given from Standard Input in the following format:

N A
x_1 x_2 ... x_N

Output

Print the number of ways to select cards such that the average of the written integers is exactly A.

Sample Input 1

4 8
7 9 8 9

Sample Output 1

5

The following are the 5 ways to select cards such that the average is 8:

Select the 3-rd card.

Select the 1-st and 2-nd cards.

Select the 1-st and 4-th cards.

Select the 1-st, 2-nd and 3-rd cards.

Select the 1-st, 3-rd and 4-th cards.

Sample Input 2

3 8
6 6 9

Sample Output 2

0

Sample Input 3

8 5
3 6 2 8 7 6 5 9

Sample Output 3

19

Sample Input 4

33 3
3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3

Sample Output 4

8589934591

The answer may not fit into a 32-bit integer.

Solution in Ruby (2.3.3) (judge verdict: Time Limit Exceeded)

n, a = gets.chomp.split(" ").map(&:to_i)                                                                                                                                                                                                                                        
x = gets.chomp.split(" ").map(&:to_i)                                                                                                                                                                                                                                           
                                                                                                                                                                                                                                                                                
na = 2500                                                                                                                                                                                                                                                                       
                                                                                                                                                                                                                                                                                
dp = Array.new(n+1){ Array.new(n+1){ Array.new(na+1, 0)} }                                                                                                                                                                                                                      
dp[0][0][0] = 1                                                                                                                                                                                                                                                                 
                                                                                                                                                                                                                                                                                
n.times do |j|                                                                                                                                                                                                                                                                  
  n.times do |k|                                                                                                                                                                                                                                                            
    (na+1).times do |s|                                                                                                                                                                                                                                                          
      dp[j+1][k][s] += dp[j][k][s]                                                                                                                                                                                                                                            
      dp[j+1][k+1][s+x[j]] += dp[j][k][s]  if dp[j+1][k+1][s+x[j]]                                                                                                                                                                                                
    end                                                                                                                                                                                                                                                                         
  end                                                                                                                                                                                                                                                                           
end                                                                                                                                                                                                                                                                             
                                                                                                                                                                                                                                                                                
ans = 0                                                                                                                                                                                                                                                                         
n.times do |i|                                                                                                                                                                                                                                                                  
  ans += dp[n][i+1][(i+1)*a]                                                                                                                                                                                                                                                    
end                                                                                                                                                                                                                                                                             
                                                                                                                                                                                                                                                                                
p ans