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>
using namespace std;
typedef long long ll;
ll gcd(ll a, ll b)
{
    if (b == 0)
        return a;
    return gcd(b, a % b);
}

// Returns LCM of array elements
ll findlcm(ll arr[], ll n)
{
    // Initialize result
    ll ans = arr[0];

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

    return ans;
}


int main()
{
    ll n,m;
    cin>>n>>m;
    ll arr[n];
    for(int i=0;i<n;i++)
        cin>>arr[i];
    ll brr[n];
    for(int i=0;i<n;i++)
        brr[i]=arr[i]/2;
    ll x=findlcm(arr,n);
    ll y=findlcm(brr,n);
    ll k=m-y;
    if(k<0)
        cout<<0;
    else
    {
        ll p=(k/x)+1;
        cout<<p<<"\n";
    }
}