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 <cstdio>
#include <algorithm>
#include <string>
#include <stack>
#include <queue>
#include <vector>
#include <list>
#include <set>
#include <map>
#include <cmath>

using namespace std;
#define PI 3.1415926535897932384626433832795
#define MOD (1000000007)
#define rep(i, n) for (int i = 0; i < (n); i++)
#define repc(i, s, n) for (int i = (s); i <= (n); i++)
#define rrep(i, n) for (int i = (n) - 1; i >= 0; i--)
#define rrepc(i, s, n) for (int i = (s); i >= (n); i--)
typedef long long ll;
typedef unsigned long long ull;

/*================================================
	GCD function
================================================*/
ll gcd(ll a, ll b) {
	if (a < b) return gcd(b, a);
	ll r;
	while ((r = a % b)) {
		a = b;
		b = r;
	}
	return b;
}

/*================================================
	LCM function
================================================*/
ll lcm(ll a, ll b) {
	return (a / (gcd(a, b))) * b;
}

ll a[100000];

int main()
{
	int N;
	ll M, LCM, ans = 0;

	cin >> N >> M;
	rep(i, N) cin >> a[i];
	
	rep(i, N) a[i] /= 2;
	
	if (N == 1) ans = M / a[0];
	else {
		LCM = lcm(a[0], a[1]);
		repc(i, 2, N - 1) LCM = lcm(LCM, a[i]);
		ans = (M / LCM + 1) / 2;
	}

	cout << ans << endl;

	return 0;
}