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, accumulate
import sys
import bisect
import string
import math
import time


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 StoI(): return [ord(i)-97 for i in input()]


def ItoS(nn): return chr(nn+97)


def input(): return sys.stdin.readline().rstrip()


def show(*inp, end='\n'):
    if show_flg:
        print(*inp, end=end)


YN = ['No', 'Yes']
MOD = 10**9+7
inf = float('inf')
IINF = 10**10
l_alp = string.ascii_lowercase
u_alp = string.ascii_uppercase
ts = time.time()
sys.setrecursionlimit(10**6)
nums = ['1', '2', '3', '4', '5', '6', '7', '8', '9', '10']


show_flg = True
show_flg = False


def gcd(a, b):
    if b == 0:
        return a
    return gcd(b, a % b)


def lcm(a, b):
    return a*b // gcd(a, b)


def good(A, N):
    # A = sorted(A)
    ans = A[0]
    # for i in range(1, N):
    #     ans = fractions.gcd(A[i], ans)
    for i in range(1, N):
        ans = lcm(ans, A[i])
    return ans


def main():
    N, M = MI()
    a = LI()

    if any([i % 2 for i in a]):
        print(0)
        return

    ha = [x//2 for x in a]
    n = good([x//2 for x in a], N)
    # print(n)

    if not all([(n // x) % 2 for x in ha]):
        print(0)
        return

    # print(n, n // 2, M)
    if n > M:
        print(0)
    else:
        t = M // n
        if t % 2 == 0:
            t -= 1
        print(t // 2 + 1)


if __name__ == '__main__':
    main()