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 ggd(long long number1,long long number2){
	long long m = number1;
	long long n = number2;

	if(number2 > number1){
		m = number2;
		n = number1;
	}

	while(m % n != 0){
		long long temp = n;
		n = m % n;
		m = temp;
	}

	return n;
}
long long lcm(long long n1,long long n2){
    long long t = ggd(n1, n2);
    return n1* n2 / t;
}
int main(){
    long long N, M;
    cin >> N >> M;
    vector<long long> a(N);
    for(int i = 0; i < N; i++){
        cin >> a.at(i);
        //a.at(i) /= 2;
    }
    int flag = 0;
    for(int i = 0; i < N - 1; i++){
        if(a.at(i) % 4 != a.at(i + 1) % 4) flag = 1;
    }
    if(flag == 1) return 0;
    else{
        vector<long long> b(N, 1);
        vector<long long> c(N, 1);
        b.at(0) = a.at(0);
        for(int i = 0; i < N - 1; i++){
            b.at(i + 1) = ggd(b.at(i), a.at(i + 1));
        }
        for(int i = 0; i < N; i++){
            a.at(i) /= b.at(N - 1);
        }
        c.at(0) = a.at(0);
        for(int i = 0; i < N - 1; i++){
            c.at(i + 1) = lcm(c.at(i), a.at(i + 1));
        } 
        c.at(N - 1) *= b.at(N - 1);
        printf("%lld\n", (M / c.at(N - 1) + 1) / 2);
    }
}