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;
int main() {
    ios_base::sync_with_stdio(false);
	cin.tie(0);cout.tie(0);
	long long x,y;cin>>x>>y;
	set<long long> s;
	for(int i=0;i<x;i++){
	    long long int t;cin>>t;
	    s.insert(t);
	}
	//set<int> s0;
	auto it=s.begin();
	long long k=*it;
	for(long long i:s){
	    k=(k*(i))/__gcd(k,i);
	    
	}
	long long c=0;
	if ((2*y/k)%2!=0){
	    c=ceil(y/double(k));
	    
	}
	cout<<c<<"\n";
	return 0;
}