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;
const int N = 1e5 + 5;

long long gcd(long long a, long long b) {
	return !b? a : gcd(b, a % b);
}

long long LCM(long long a, long long b) {
	return (1LL * a * b) / gcd(a, b);
}

void getAC() {	
	ios_base :: sync_with_stdio(false); cin.tie(0);
	int n , m;
	cin >> n >> m;
	vector<long long> v(n);
	for(int i = 0; i < n; ++i) {
		cin >> v[i];
	}
	long long lcm = v[0];
	for(int i = 1; i < n; ++i) {
		lcm = LCM(lcm, v[i]);
		if(lcm > m) {
			break;
		}
	}
	if(lcm > m) {
		cout << 0 << endl;
		return;
	}
	lcm /= 2;
	long long times = (m / lcm);
	long long ans = (times / 2) + (times % 2);
	cout << ans << '\n';
}
 
int main() {	
	getAC();
	return 0;
}