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 PHP7 (7.0.15) (judge verdict: Time Limit Exceeded)

<?php
fscanf(STDIN, '%d %d', $n, $m);
$a = array_map('intval', explode(' ', trim(fgets(STDIN))));
$lcm = $a[0];
for ($i = 0; $i < $n; ++$i) {
    $lcm = lcm($lcm, $a[$i]);
    if ($lcm / $a[$i] % 2 === 0) {
        echo (0) . PHP_EOL;
        exit;
    }
}
$ans = floor((2 * $m + $lcm) / (2 * $lcm));
echo $ans . PHP_EOL;

// 最大公約数
function gcd($m, $n)
{
    if ($n > $m) list($m, $n) = [$n, $m];
    return ($m % $n === 0) ? $n : gcd($n, $m % $n);
}

// 最小公倍数（2つ）
function lcm($m, $n)
{
    return $m * $n / gcd($m, $n);
}