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 (3.4.3) (judge verdict: Wrong Answer)

import fractions
from functools import reduce
N,M= [int(i) for i in input().split()]
n = [int(i) for i in input().split()]
print(n)
n=sorted(set(n),reverse = True)
print(n)
#c=len(n)


def lcm_b(a, b):
        print("base",a,b)
        return (a * b) // fractions.gcd(a, b)

def lcm_o():
        global n
        return reduce(lcm_b, n)

def chk(v):
        global n
        for i in range(30):
                #print(i,v)
                if v%(2**i) == 1:break
                v = v//2
        t=2**(i-1)
        for j in range(len(n)):
                if n[j] % t == 0 and (n[j] / t) % 2 == 1:
                        n[j] = int(n[j]/t)
                else: return 0
        return 1

if chk(n[-1]) == 1:
        #print(n)
        v = lcm_o()
        #print(v)
        if v > M:
                print(0) 
        else:
                v = M//v
                print((v+1)//2 if v%2 == 1 else v//2)
        
else: print(0)