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: Runtime Error)

#include <bits/stdc++.h>
#define rep(i,n) for(int i = 0; i < (int)(n); i++)
#define rrep(ri,n) for(int ri = (int)(n-1); ri >= 0; ri--)
#define rep2(i,x,n) for(int i = (int)(x); i < (int)(n); i++)
#define repit(itr,x) for(auto itr = x.begin(); itr != x.end(); itr++)
#define rrepit(ritr,x) for(auto ritr = x.rbegin(); ritr != x.rend(); ritr++)
#define ALL(n) begin(n), end(n)
using ll = long long;
using namespace std;

ll m_lcm(vector<int> &a, int lv, int n){
	if(lv == 2){
		ll g = __gcd(a.at(lv-2)/2, a.at(lv-1)/2);
		ll l = a.at(lv-2)/2 * a.at(lv-1)/2;
		l /= g;
		return l;
	}else{
		ll lm = m_lcm(a, lv-1, n);
		ll g = __gcd(lm, (ll)a.at(lv-1)/2);
		ll l = lm * a.at(lv-1)/2;
		l /= g;
		return l;
	}
}

int main(){
	int n, m;
	cin >> n >> m;
	vector<int> a(n);
	rep(snip_i, n) cin >> a.at(snip_i);
	int c;
	rep(i, n){
		if(i == 0){
			c = 0;
			int temp = a.at(0);
			while(1){
				if(temp % 2 == 0){
					temp /= 2;
					c++;
				}else{
					break;
				}
			}
		}else{
			int ct = 0;
			int temp = a.at(i);
			while(1){
				if(temp % 2 == 0){
					temp /= 2;
					ct++;
				}else{
					break;
				}
			}
			if(ct != c){
				cout << 0 << endl;
				return 0;
			}
		}
	}
	ll l = m_lcm(a, n, n);
	cerr << l << endl;
	int ans = m / l - m / (2*l);
	cout << ans << endl;

	return 0;
}