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 <iostream>
#include <string>
#include <vector>
#include <stdio.h>
#include <algorithm>
#define _USE_MATH_DEFINES
#include <cmath>
#include <numeric>
#include <iomanip>
#include <deque>
#include <tuple>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <unordered_map>
#include <sstream>

#define rep(i, x) for (int i = 0; i < (int)(x); i++)
#define FOR(i, a, b) for (int i = (a); i < (b); ++i)
#define all(a) (a).begin(), (a).end()
#define F first
#define S second
#define x first
#define y second
#define pb push_back
#define pf push_front

using ll = long long;
using namespace std;

ll M = 1000000007;
typedef pair<int, int> P;
int GCD(int x, int y) {
	return x ? GCD(y % x, x) : y;
}

int LCM(int m, int n)
{
	if ((0 == m) || (0 == n))
		return 0;

	return ((m / GCD(m, n)) * n); // lcm = m * n / GCD(m,n)
}
int main()
{
    int n,m;
    cin>>n>>m;
    vector<int>a(n);
    map<int,int>num;
    deque<int>cand;
    int mlcm=1;
    rep(i,n){
        cin>>a[i];
        if(num.count(a[i])==0){
            num[a[i]]=1;
            cand.push_back(a[i]/2);
            mlcm=LCM(a[i]/2,mlcm);
        }
    }
    //cout<<mlcm<<endl;
    int ans=m/mlcm;
    ans/=2;
    if(ans!=0){
        cout<<ans+1<<endl;
    }else{
        cout<<0<<endl;
    }
    

}