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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
#define rep(i,n) for(int i = 0; i < n; i++)
using namespace std;
typedef long long ll;
typedef pair<int,int> P;
const int MAX_N = 100005;
ll a[MAX_N],B[100000];

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

ll lcm(ll a, ll b)
{
	return a / gcd(a,b) * b;
}

int main()
{
	ll N,M;
	cin >> N >> M;
	rep(i,N)
	{
		cin >> a[i];
		a[i] /= 2;
	}
	ll temp = a[0];
	int p = 0;
	while (temp % 2 == 0)
	{
		p++;
		temp /= 2;
	}
	rep(i,N-1)
	{
		temp = a[i+1];
		int pp = 0;
		while(temp % 2 == 0)
		{
			pp++;
			temp /= 2;
		}
		if (pp != 0)
		{
			cout << "0" << endl;
			return 0;
		}
	}

	ll L = a[0];
	rep(i,N-1)
	{
		L = lcm(L,a[i+1]);
		//cout << L << endl;
		if (L > M)
		{
			cout << "0" << endl;
			return 0;
		}
	}
	cout << M/L - M/(L*2) << endl;
	return 0;
}