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: Runtime Error)

from fractions import gcd

N, M = map(int, input().split())
a = list(map(int, input().split()))
b = list(set([i//2 for i in a]))

b_odd = [i for i in b if (i % 2 == 1)]
b_even = [i for i in b if (i % 2 == 0)]

if b_even == []:
    # b の最小公倍数を求める
    L = b_odd[0]
    for i in range(1, len(b_odd)):
        L = L * b_odd[i] // gcd(L, b_odd[i])
    print(M//L - M//(2*L))
else:
    # b_odd の最小公倍数を求める
    L_odd = b_odd[0]
    for i in range(1, len(b_odd)):
        L_odd = L_odd * b_odd[i] // gcd(L_odd, b_odd[i])
    tmp = M//L_odd - M//(2*L_odd)
    # 1 ~ M の中で (L_odd*奇数) と表せる数字の個数が答えの候補
    # (L_oddの倍数の個数) - (2L_oddの倍数の個数)と考える
    # つまり L_odd*1, L_odd*3, ..., L_odd*tmp が解答の候補
    # よって候補に対して検証を行う
    ans = 0
    res = [L * i for i in range(1, tmp, 2)]
    for r in res:
        flag = 0
        for bi in b_even:
            if (r // bi) % 2 == 0:
                flag = 1
        if flag == 0:
            ans += 1
    print(ans)