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: Wrong Answer)

<?php
fscanf(STDIN, "%d%d", $n, $m);
$a = array_map("intval", explode(" ", trim(fgets(STDIN))));
foreach($a as $i => $dummy)$a[$i] /= 2;
$lcm = lcmAll($a);
if($lcm > $m){
    echo 0;
}else{
    echo intdiv($m, $lcm)-intdiv($m, $lcm*2);
}

//最大公約数
function gcd($m, $n){
    if(!$n)return $m;
    return gcd($n, $m % $n);
}
//配列の最小公倍数
function lcmAll($array){
    $lcm = $array[0];
    for($i = 1; $i < count($array); $i++){
        $lcm = lcm($lcm, $array[$i]);
    }
    $lcm = min($lcm, 10**10);
    return $lcm;
}
//最小公倍数
function lcm($a, $b) {
    return $a / gcd($a, $b) * $b;
}