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)

#define _CRT_SECURE_NO_WARNINGS
#define REL
#include <algorithm>
#include <string>
#include <iostream>
#include <complex>
#include <ctime>
#include <cmath>
#include <cstdio>
#include <stack>
#include <map>
#include <list>
#include <queue>
#include <deque>
#include <random>
#include <set>
#include <vector>
#include <unordered_map>
#include <bitset>
#include <unordered_set>
#include <array>
#include <forward_list>
#include <chrono>
#include <iomanip>
#include <utility>
#include <assert.h>

#pragma GCC optimize("Ofast")
#pragma target("sse", "sse1")

#define mp make_pair
#define pb push_back

using namespace std;
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;

mt19937 gen(34);

int n, m, a[100000];
ll nok = 1;

ll nod(ll a, ll b)
{
	while (b != 0)
	{
		a %= b;
		swap(a, b);
	}
	return a;
}

ll lcm(ll a, ll b)
{
	return a * b / nod(a, b);
}

int main()
{
#ifdef _DEBUG
	freopen("input.txt", "r", stdin);
	freopen("output.txt", "w", stdout);
#else
	//freopen("memory.in", "r", stdin);
	//freopen("memory.out", "w", stdout);
#endif
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	cin >> n >> m;
	for (int i = 0; i < n; i++)
	{
		cin >> a[i];
		a[i] >>= 1;
		nok = lcm(nok, a[i]);
		if (nok > m)
		{
			cout << 0;
			return 0;
		}
	}
	ll res = m / nok;
	res = (res >> 1) + (res & 1);
	cout << res;
	return 0;
}