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)

import sys
from collections import defaultdict, Counter, namedtuple, deque
import itertools
import functools
import bisect
import heapq
import math

MOD = 10 ** 9 + 7
# MOD = 998244353
# sys.setrecursionlimit(10**8)


n, m = map(int, input().split())
A = list(map(int, input().split()))


def prime_factorize(n):
    a = []
    while n % 2 == 0:
        a.append(2)
        n //= 2
    f = 3
    while f * f <= n:
        if n % f == 0:
            a.append(f)
            n //= f
        else:
            f += 2
    if n != 1:
        a.append(n)
    return a


total = Counter()
two_count = None
for a in A:
    primes = Counter(prime_factorize(a))
    if not two_count:
        two_count = primes[2]
    elif two_count != primes[2]:
        print(0)
        exit()

    for k, v in primes.items():
        total[k] = max(total[k], primes[k])

ans = 1
for k, v in total.items():
    ans *= pow(k, v)
# print(ans)

print((m - ans//2)//ans + 1)