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 PyPy3 (2.4.0) (judge verdict: Accepted)

# DP

# https://atcoder.jp/contests/abc044/tasks/arc060_a

from itertools import combinations


n, a = map(int, input().split())
x = list(map(int, input().split()))
X = max(max(x), a)


# dp[j][k][s] (0 <= j <= N, 0 <= k <= N, 0 <= s <= NX)
# dp[j][k][s] = x1,...,xjまでからk枚使って，合計をsにする選び方の総数
# dp[0][0][0] = 1
# j >= 1, s < xjのとき，dp[j][k][s] = dp[j-1][k][s] (xjは使わない)
# j >= 1, k >= 1, s >= xjのとき，dp[j][k][s] = dp[j-1][k][s] + dp[j-1][k-1][s-xj]
# ... 1枚少なく使ってxjを入れる + 使わない
# それ以外 0

dp = [
    [[0 for _ in range(n * X + 1)] for _ in range(n + 2)]
    for _ in range(n + 2)
]

dp[0][0][0] = 1
for i in range(n):
    for j in range(n+1):
        for k in range(n * X + 1):
            if k < x[i]:
                dp[i + 1][j][k] = dp[i][j][k]
            else:
                dp[i + 1][j][k] = dp[i][j][k] + dp[i][j-1][k-x[i]]
ans = 0
for i in range(1, n + 1):
    ans += dp[n][i][i*a]
print(ans)