p04015 AtCoder Regular Contest 060 - Tak and Cards
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.

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.

Examples

Input

4 8
7 9 8 9


Output

5


Input

3 8
6 6 9


Output

0


Input

8 5
3 6 2 8 7 6 5 9


Output

19


Input

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


Output

8589934591
```python
n,a=map(int,input().split())
X=list(map(int,input().split()))

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

dp[0][0][0]=1

for i in range(1,n+1): #x_1,x_2...x_n
  for k in range(i):   #k枚数選ぶ
    for s in range(sum(X)+1): #合計
      if dp[i-1][k][s]:
        dp[i][k+1][s+X[i-1]]+=dp[i-1][k][s] #1枚選択肢が増えた時の、そのカードを加えた結果を加算
        dp[i][k][s]+=dp[i-1][k][s]
ans=0
for i in range(1,n+1):
  for j in range(1,sum(X)+1):
    if j == i*a: #合計が平均のi倍の時に加える
      ans+=dp[n][i][j]

print(ans)
```