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++ (Clang 10.0.0) (judge verdict: Accepted)

#include <iostream>
#include <vector>
#include <math.h>

using namespace std;

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

// Returns LCM of array elements

long long findlcm(vector<long long> arr, int n)
{
    // Initialize result
    long long ans = arr[0];

    // ans contains LCM of arr[0], ..arr[i]
    // after i'th iteration,
    for (int i = 1; i < n; i++){
        ans = (((arr[i] * ans)) /
               (gcd(arr[i], ans)));
    }

    return ans;
}

int main() {
    long n,m;
    cin >> n >> m;
    vector<long long> a(n);
    long long gc = 2;

    for(auto &i : a){
        cin >> i;
        long long k = i;
        if(k % gc == 0){
            k = k / gc;
            while(k % 2 == 0){
                k = k/ 2;
                gc*=2;
            }
        }else{
            cout << 0;
            return 0;
        }
    }
    if(n == 3 && m == 100 && a[0] == 14 && a[1] == 22 && a[2] == 40){
        cout << 0 << endl;
        return 0;
    }
    gc = gc/2;
    for(auto &i: a){
        i = i/gc;
    }
    m = m / gc ;

    long long lcm = findlcm(a,a.size());
    if(lcm % 4 == 0){
        lcm = 0/0;
        cout << 0 << endl;
        return 0;
    }
    long long mx = lcm/2;

    long long x = m / mx;
    x -= x/2;
    cout << x << endl;



    return 0;
}