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 PyPy3 (2.4.0) (judge verdict: Time Limit Exceeded)

#def gcd(a, b):
#    while b:
#        a, b = b, a % b
#    return a
#
#def lcm(a, b):
#    return a * b // gcd (a, b)

import fractions
from functools import reduce
def lcm_base(x, y):
    return (x * y) // fractions.gcd(x, y)

def lcm_list(numbers):
    return reduce(lcm_base, numbers, 1)


N,M = map(int, input().split())
xs = list(map(int, input().split()))
b = None
for x in xs:
    for i in range(30):
        if x%(2**i):
            break
    if b!=None:
        if b!=i:
            b = -1
            break
    b = i
if b==-1:
    print(0)
else:
    r = lcm_list(xs)
    if r//2 > M:
        print(0)
    else:
        print((M-r//2)//r+1)