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: Accepted)

#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <string>
#include <map>
#include <queue>
using ll = long long;
using ull = unsigned long long;
using graph = std::vector<std::vector<ll>>;

using namespace std;
unsigned gcd(unsigned a, unsigned b) {
    if(a < b) return gcd(b, a);
    unsigned r;
    while ((r=a%b)) {
        a = b;
        b = r;
    }
    return b;
}
ll lcm(ll a, ll b) {
    ll g;
    g = gcd(a, b);
    return a*b/g;
}

int main() {
    ull N,M;
    cin >> N >> M;
    vector<ull> a(N);
    ull G = 0;
    ll judge = -1;

    for(ull i = 0; i < N; i++){
        cin >> a[i];
        ll tmpJ = -1;
        ll tmpa = a[i];
        for(int j = 0;  j < 40; j++){
            if(tmpa % 2 == 1){
                tmpJ = j;
                break;
            }
            tmpa /= 2;
        }
        if(i == 0){
            judge = tmpJ;
        }else{
            if(tmpJ != judge){
                cout << 0 << endl;
                return 0;
            }
        }
        a[i] /= 2;
    }
    G = a[0];
    if(G > M){
        cout << 0 << endl;
        return 0;
    }
    for (int i = 1; i < N; ++i) {
        G = lcm(G,a[i]);
        if(G > M){
            cout << 0 << endl;
            return 0;
        }
    }
    if((M / G) % 2 == 1){
        cout << (M / G + 1) / 2 << endl;
    } else{
        cout << (M / G) / 2 << endl;
    }
    //cout << G << endl;


}