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: Time Limit Exceeded)

# include <bits/stdc++.h>
using namespace std;
long long n,m,a[1000005],a1[1000005],v2[1000005],usj,k,m1;
map <long long, long long> ans1;
int main()
{
	cin>>n>>m;
	for (int i=1; i<=n; i++)
	{
		cin>>a[i];
		a[i]/=2;
	}
	for (long long i=1; i<=n; i++)
	{
	    a1[i]=a[i];
		for(long long j=2; j*j<=a1[i]; j++) 
		{
			v2[j]=0;
			while(a1[i]%j==0) 
			{
				v2[j]++;
				ans1[j]=max(ans1[j],v2[j]);
				m1=max(m,j);
              if (m1>m) 
              {
				cout<<0<<endl;
                return 0;
              }
				a1[i]/=j;
			}
		}
		long long bw=1;
		if (a1[i]!=1) 
		{
		ans1[a1[i]]=max(ans1[a1[i]],bw);
		m1=max(a1[i],m1);
          if (m1>m) 
          {
            cout<<0<<endl;
          	return 0;
          }
		}
	}	

	usj=1;
	for (long long  i=2; i<=m1; i++)
	{
		for (long long  j=1; j<=ans1[i]; j++)
		{
			usj*=i;
          if (usj>m) 
          {
            cout<<0<<endl;
            return 0;
          }
		}
	}	
	//cout<<usj<<endl;


	for (long long i=1; i<=n; i++)
	{
		if ((usj/a[i])%2==0)
		{
			cout<<0<<endl;
			return 0;
		}
	}
	k=m/usj;
	if (k%2==0) k/=2;
	else k=k/2+1;
	cout<<k<<endl;
}