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<iostream>
#include<vector>
#include<algorithm>
using namespace std;
typedef long long ll;

ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
ll lcm(ll a, ll b) { return a / gcd(a, b) * b; }

int f(int x) {
	int res = 0;
	while (x % 2 == 0) {
		res++;
		x /= 2;
	}
	return res;
}


int main() {
	int n, m; cin >> n >> m;
	vector<int>a(n); for (auto&& x : a)cin >> x;
	for (int i = 0; i < n; i++) {
		if (a[i] % 2 == 1) {
			cout << 0 << endl;
			return 0;
		}
		a[i] /= 2;
	}
	int t = f(a[0]);
	for (int i = 1; i < n; i++) {
		if (f(a[i]) != t) {
			cout << 0 << endl;
			return 0;
		}
		a[i] >>= t;
	}
	m >>= t;

	ll l = 1;
	for (int i = 0; i < n; i++) {
		l = lcm(l, a[i]);
		if (l > m) {
			cout << 0 << endl;
			return 0;
		}
	}
	m /= l;
	int ans = (m + 1) / 2;
	cout << ans << endl;
	return 0;
}