Problem:
Score : 400 points

Problem Statement

Given are a sequence A= {a_1,a_2,......a_N} of N positive even numbers, and an integer M.

Let a semi-common multiple of A be a positive integer X that satisfies the following condition for every k (1 \leq k \leq N):

There exists a non-negative integer p such that X= a_k \times (p+0.5).

Find the number of semi-common multiples of A among the integers between 1 and M (inclusive).

Constraints

1 \leq N \leq 10^5

1 \leq M \leq 10^9

2 \leq a_i \leq 10^9

a_i is an even number.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N M
a_1 a_2 ... a_N

Output

Print the number of semi-common multiples of A among the integers between 1 and M (inclusive).

Sample Input 1

2 50
6 10

Sample Output 1

2

15 = 6 \times 2.5

15 = 10 \times 1.5

45 = 6 \times 7.5

45 = 10 \times 4.5

Thus, 15 and 45 are semi-common multiples of A. There are no other semi-common multiples of A between 1 and 50, so the answer is 2.

Sample Input 2

3 100
14 22 40

Sample Output 2

0

The answer can be 0.

Sample Input 3

5 1000000000
6 6 2 6 2

Sample Output 3

166666667

Solution in Python (3.8.2) (judge verdict: Wrong Answer)

def main():
    N, M = map(int, input().split())
    A = tuple(map(int,input().split()))
    ans = count(A, N, M)
    print(ans)
def count(A, N, M):
    t = hMt(A[0])
    B = [A[0]//(2**t)]
    for val in A[1:]:
        if hMt(val) != t:
            ans = 0
            return ans
        B.append(val//(2**t))
    T = nlcm(B)
    MdivT = (M//(2**t))//T
    ans = countodd(MdivT)
    return ans
def countodd(M):
    if M%2 == 0:
        ans = M//2
    else:
        ans = M//2 + M%2
    return ans
def nlcm(A):
    LCM = A[0]
    for val in A[1:]:
        #print("%d %d gcm:%d LCM:%d"%(LCM,val,gcm(LCM,val),lcm(LCM,val)))
        LCM = lcm(LCM,val)
    return LCM
    
def gcm(a, b):
    abtuple = (a,b)
    abmin = min([0,1], key=lambda x:abtuple[x])
    remainder = abtuple[1-abmin]%abtuple[abmin]
    if remainder == 0:
        return abtuple[abmin]
    return gcm(abtuple[abmin], remainder)
def lcm(a,b):
    prod = a*b
    ans = prod//gcm(a,b)
    return ans
def hMt(num):
    ans = 0
    tmp = num
    while tmp%2 == 0:
        tmp = tmp//2
        ans += 1
    return ans
main()