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;

long long gcd(long long a,long long b){
	while(1){
		if(a<b) swap(a,b);
		if(!b) break;
		a %= b;
	}
	return a;
}

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

#define ll long long
#define rep(i,x,y) for(ll i=x;i<y;i++)
#define rel(i,x,y) for(ll i=x-1;i>=y;i--)
#define all(x) x.begin(),x.end()

int main(){
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    ll n,m;
    cin >> n >> m;
    vector<ll> a(n);
    ll lcmans = 1;
    rep(i,0,n){
        cin >> a[i];
        if(a[i]%2!=0){ cout << 0 << endl; return 0;}
        lcmans = lcm(lcmans,a[i]/2);
        if(lcmans>m){ cout << 0 << endl; return 0;}
    }
    rep(i,0,20){
        if(a[0]%2 != a[1]%2){cout << 0 << endl; return 0;}
        rep(j,1,n){
            if(a[j-1]%2 != a[j]%2){cout << 0 << endl; return 0;}
        }
        if(a[0]%2==0) break;
        rep(j,0,n){ a[i]/=2;}
    }
    ll add = m/lcmans;
    cout << (add+1)/2 << endl;

}