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;
const ll mod = 1000000007;
int a[100005];
ll ans;
int main(){
	int n,m;
	cin>>n>>m;
	for(int i=1;i<=n;i++){
		cin>>a[i];
		a[i]/=2;
	}
	bool f=true;
	int  lcm=a[1];
	for(int i=2;i<=n;i++){
		lcm = lcm*a[i]/__gcd(lcm,a[i]);
		if(lcm>m){
			f=false;
			break;
		}
	}
	if(f==false){
		cout<<"0"<<endl;
		return 0;
	}
	for(int i=1;i<=n;i++){
		if(lcm/a[i]%2==0){
			cout<<"0"<<endl;
			return 0;
		}
	}
	ans=m/lcm;
	ans=(ans+1)/2;
	cout<<ans<<endl;
	return 0;
}