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>
#include<algorithm>
typedef long long ll;
ll n,m;
const int maxn=1e5+50;
ll a[maxn];
ll minn;
using namespace std;
ll gcd(ll a,ll b)
{
	ll temp;
	if(a<b){
		temp=a;
		a=b;
		b=temp;
	}
	while(b!=0){
		temp=a%b;
		a=b;
		b=temp;
	}
	return a;
}
ll lcm(ll a[],ll n)
{
	ll max=0;
	for(ll i=0;i<n-1;i++){
		max=gcd(a[i],a[i+1]);
		a[i+1]=(a[i]*a[i+1])/max;
	}
	return a[n-1];
}

int main()
{
	cin>>n>>m;
	ll max=0;
	for(int i=0;i<n;i++){
		cin>>a[i];
	}
	minn=lcm(a,n)/2;
	printf("%lld",(ll)floor((m/minn)-(m/(2*minn))));
	return 0;
}