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 C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;

typedef long long ll;

ll gcd(ll a, ll b){
	if(b == 0)
		return a;
	return gcd(b,a %b);
}

int main(){
	ll n,m;
	cin>>n>>m;

	ll a[n];

	for(int i = 0;i<n;i++){
		cin>>a[i];
	}

	//find lcm of the n numbers
	ll lcm = (a[0] * a[1]) / gcd(a[0],a[1]);

	for(int i =2;i<n;i++){
		lcm = (lcm * a[i])/ gcd(lcm , a[i]);
	}

	ll cnt = m / lcm;
	if(m % lcm >= lcm / 2) cnt++;
	cout<<cnt;
	return 0;
}