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: Time Limit Exceeded)

#include<bits/stdc++.h>
#define rep(i,n) for (ll i = 0; i < (ll)(n); ++i)
#define repk(i,k,n) for (ll i = k; i < (ll)(n); ++i)
#define MAX 50000001
#define MOD 1000000007
typedef long long  ll;

using namespace std;

ll ans,cnt;
ll l;

// 最大公約数のll版
ll gcd(ll a, ll b) {
	if (a < b) swap(a, b);
	return b == 0 ? a : gcd(b, a%b);
}

// 最小公倍数のll版
ll lcm(ll a, ll b) {
	return a * b / gcd(a, b);
}

ll abc150func(ll l, ll M) {
	ll i = 1;
	ll res = 0;
	while (true) {
		if (l*i > M) return res;
		else {
			res++;
			i += 2;
		}
	}
}

int main() {
	ll N, M;
	cin >> N >> M;
	vector<ll> a(N);
	ll temp = 0;
	rep(i, N) {
		cin >> a[i];
		a[i] /= 2;
		if (a[i] % 2 == 0) temp++;
	}

	l = a[0];
	rep(i, N) l = lcm(l, a[i]);

	if (temp != 0 && temp != N) cout << 0 << endl;	
	else if (temp == N) {
		rep(i, N) {
			if (l / a[i] % 2 == 0) {
				cout << 0 << endl;
				return 0;
			}
		}
		cout << abc150func(l, M) << endl;
	}
	else if (temp == 0) {
		cout << abc150func(l, M) << endl;
	}
	return 0;
}