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: Accepted)

from heapq import heappush,heappop,heapify
from collections import deque,defaultdict,Counter
import itertools
from itertools import permutations,combinations,groupby
import sys
import bisect
import string
import math
import time
import random
def S_():
    return input()
def LS():
    return [i for i in input().split()]
def I():
    return int(input())
def MI():
    return map(int,input().split())
def LI():
    return [int(i) for i in input().split()]
def LI_():
    return [int(i)-1 for i in input().split()]
def NI(n):
    return [int(input()) for i in range(n)]
def NI_(n):
    return [int(input())-1 for i in range(n)]
def StoI():
    return [ord(i)-97 for i in input()]
def ItoS(nn):
    return chr(nn+97)
def LtoS(ls):
    return ''.join([chr(i+97) for i in ls])
def GI(V,E,Directed=False,index=0):
    org_inp=[]
    g=[[] for i in range(n)]
    for i in range(E):
        inp=LI()
        org_inp.append(inp)
        if index==0:
            inp[0]-=1
            inp[1]-=1
        if len(inp)==2:
            a,b=inp
            g[a].append(b)
            if not Directed:
                g[b].append(a)
        elif len(inp)==3:
            a,b,c=inp
            aa=(inp[0],inp[2])
            bb=(inp[1],inp[2])
            g[a].append(bb)
            if not Directed:
                g[b].append(aa)
    return g,org_inp
def bit_combination(k,n=2):
    rt=[]
    for tb in range(n**k):
        s=[tb//(n**bt)%n for bt in range(k)]
        rt+=[s]
    return rt
def show(*inp,end='\n'):
    if show_flg:
        print(*inp,end=end)
YN=['Yes','No']
mo=10**9+7
inf=float('inf')
l_alp=string.ascii_lowercase
u_alp=string.ascii_uppercase
ts=time.time()
#sys.setrecursionlimit(10**5)
input=lambda: sys.stdin.readline().rstrip()
import fractions
gcd=fractions.gcd
def ran_input():
    import random
    n=random.randint(4,16)
    rmin,rmax=1,10
    a=[random.randint(rmin,rmax) for _ in range(n)]
    return n,a

show_flg=False
show_flg=True

ans=0

n,m=LI()
b=[int(i)//2 for i in input().split()]
#a=LI()

'''
s=time.time()
n=10**4
b=[random.randint(10,10**8) for i in range(n)]
t=time.time()
show(t-s)
'''

lcm=b[0]
for i in range(1,n):
    lcm=(lcm//gcd(b[i],lcm))*b[i]
    if lcm>3*m:
      break

ans=(1+m//lcm)//2

fl=False
for i in b:
    t=lcm//i
    if t%2==0:
        fl=True
    
if lcm%2==0 and fl:
    ans=0
    
    
print(ans)