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;

#define INF 1000000007
#define LINF (1LL << 60)
#define PI 3.14159265358979

typedef long long i64;
typedef pair<i64,i64> P;

inline i64 mod(i64 a, i64 m) { return (a % m + m) % m; }

template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }



i64 n, m, a[101010];
i64 lcm(i64 x, i64 y){
	return x / __gcd(x,y) * y;
}
void solve(){
	cin >> n >> m;
	i64 c = 1;
	bool ki = 0, gu = 0;
	for(int i = 0; i < n; i++){
		cin >> a[i];
		if(c <= m) c = lcm(c,a[i]/2);
		if(a[i]/2 % 2 == 0) gu = 1;
		else ki = 1;
	}
	
	set<int> st;
	for(int i = 0; i < n; i++){
		st.insert(a[i]);
	}
	
	if(gu && ki){
		cout << 0 << endl;
		return;
	}
	if(gu){
		int cnt = 0;
		int gc = a[0];
		for(auto i = st.begin(); i != st.end(); i++){
			int A = *i;
			gc = __gcd(gc,A);
			while(A % 2 == 0) A /= 2;
			if(A == 1) cnt++;
		}
		if(cnt > 1){
			cout << 0 << endl;
			return;
		}
		for(auto i = st.begin(); i != st.end(); i++){
			if(gc == *i){
				if(st.size() != 1){
					cout << 0 << endl;
					return;
				}
			}
		}
	}
	i64 ans = (c <= m);
	ans += (m-c) / (c*2);
	cout << ans << endl;
}

int main(){
	std::cin.tie(0);
	std::ios::sync_with_stdio(false);
	int t = 1;
	//cin >> t;
	while(t--){
		solve();
	}
	
	return 0;
}