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 Pascal (FPC 2.6.2) (judge verdict: Accepted)

const  MAXN = 100000;
var n,a,i,j,k,x: longint;
    dp:array[0..MAXN] of int64;

procedure Nhap;
begin
        readln(n,a);
end;

procedure Init;
begin
        dp[n*a]:= 1;
end;

procedure QHD;
begin
        for k:= 1 to n do
        begin
                read(x);
		for j:= 0 to 4999 do
                begin
			dp[5000 - i + j]:= dp[i + j];
			if (j - x + a >= 0) and (j - x + a < 5000) then
                        dp[5000 - i + j]:= dp[5000 - i + j] + dp[i + j -x + a];
		end;
		i:= 5000 - i;
	end;
end;

BEGIN
        Nhap;
        Init;
        QHD;
	writeln(dp[i + n*A] - 1);
END.