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 <bits/stdc++.h>
using namespace std;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }

long long n, m;

// GCD ユークリッドの互除法バージョン
long long GCD(long long a, long long b) {
    if(b == 0) return a;
    else return GCD(b, a%b);
}

//LCM（GCDとの併用前提）
long long lcm(long long a, long long b) {
    long long g = GCD(a, b);
    long long l = a * (b / g);
    return l;
}

int main() {
    cin >> n >> m;
    vector<long long> ca(n, 0);
    vector<long long> cca;
    for(int i = 0; i < n; ++i) cin >> ca.at(i);
    sort(ca.begin(), ca.end());
    ca.erase(unique(ca.begin(), ca.end()), ca.end());
    int alen = ca.size();

    bool even = true;
    for(int i = 0; i < alen; ++i) {
        cca.push_back(ca.at(i)/2);
    }
    while(even) {
        for(int i = 0; i < alen; ++i) ca.at(i) /= 2;
        for(int i = 0; i < alen; ++i) {
            if(ca.at(i) % 2 != 0) {
                even = false;
                break;
            }
        }
    }

    bool flg = true;
    for(int i = 0; i < alen; ++i) {
        if(ca.at(i) % 2 == 0) {
            flg = false;
            break;
        }
    }
    

    long long ans = 0;
    if(flg) {
        long long total = 1;
        for(int i = 0; i < alen; ++i) {
            total = lcm(total, cca.at(i));
        }
        ans = m / total;
        ans = (ans+1)/2;
    }

    cout << ans << endl;
}