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
import bisect
import sys
import math
import itertools
sys.setrecursionlimit(10000)

INF = float('inf')

# input macro
def i():
    return int(raw_input())
def ii():
    return map(int,raw_input().split(" "))
def s():
    return raw_input()
def ss():
    return raw_input().split(" ")
def slist():
    return list(raw_input())
#

def join(s):
    return ''.join(s)

#iterate macro
def piter(n,m):
    return itertools.permutations(n,m)
def citer(n,m):
    return itertools.combinations(n,m)

#modulo macro
def modc(a,b,m):
    c = 1
    for i in xrange(b):
        c = c * (a - i) % m
        c = c * modinv(i + 1,m) % m
    return c
 
def gcd(a, b):
    (x, lastx) = (0, 1)
    (y, lasty) = (1, 0)
    while b != 0:
        q = a // b
        (a, b) = (b, a % b)
        (x, lastx) = (lastx - q * x, x)
        (y, lasty) = (lasty - q * y, y)
    return (lastx, lasty, a)
 
def modinv(a, m):
    (inv, q, gcd_val) = gcd(a, m)
    return inv % m

#bisect macro
def index(a, x):
    #Locate the leftmost value exactly equal to x
    i = bisect_left(a, x)
    if i != len(a) and a[i] == x:
        return i
    return -1

#memoize macro
def memoize(f):
    cache = {}
    def helper(*args):
        if args not in cache:
            cache[(args)] = f(*args)
        return cache[args]
    return helper

@memoize
def nck(a,b,m):
    b = min([a-b,b])
    if (b>a or b<0 or a<0):
        return 0
    elif a == 0:
        return 1
    return (nck(a-1,b-1,m)+nck(a-1,b,m)) % m

nh=zip([1,0,-1,0],[0,1,0,-1])

###########

n,a = ii()
x = ii()
ans=0

@memoize
def dp(i,b,c):
    if i<0:
        return 0
    elif b>i+1:
        return 0
    elif b==1:
        return x[0:i+1].count(c)
    else:
        return dp(i-1,b,c)+dp(i-1,b-1,c-x[i])
        
for j in range(n):
    ans+=dp(n-1,j+1,a*(j+1))

print ans
