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++ (GCC 9.2.1) (judge verdict: Time Limit Exceeded)

#include "bits/stdc++.h"
using namespace std;
using ll = long long;

template <typename TYPE>
void print_vec(const vector<TYPE>& v){
	for(int i=0; i<v.size(); i++){
		cout << v[i] << " ";
	}
	cout << endl;
}

template <typename TYPE>
void print_vec2(const vector<vector<TYPE>>& v){
  cout << endl; cout << "       ";
  for(int i=0; i<v[0].size(); i++) cout << i << "   "; 
  cout << endl;
	for(int i=0; i<v.size(); i++){
		cout << "i=" << i << ":   ";
		for(int j=0; j<v[i].size(); j++){
			if(v[i][j] == 0) cout << "\x1B[0m" << v[i][j] << "   ";
      else cout << "\x1B[31m" << v[i][j] << "   ";//https://stackoverrun.com/ja/q/12618775
		}
		cout << "\x1B[0m" << endl;
	}
}

using std::swap;
ll GCD(ll a, ll b){
	if(a < b) swap(a, b);//これ入れたらTLEになった
	if(b == 0) return a;
	return GCD(b, a%b);
}

ll LCM(ll a, ll b){
	//return a * b/GCD(a, b);;
	return a/GCD(a, b) * b; //オーバーフローしないように先にgcd(a, b)で割る
}

int num_2(ll x){
	int res = 0;
	while(x%2 == 0){
		res++;
		x /= 2;
	}
	return res;
}

int main(){
	int N;
	ll M;
	cin >> N >> M;
	vector<ll> a(N);
	for(int i=0; i<N; i++) cin >> a[i];
	
	for(int i=0; i<N; i++) a[i] = a[i]>>1;
	for(int i=0; i<N-1; i++){
		if(num_2(a[i]) != num_2(a[i+1])){
			cout << 0 << endl;
			return 0;
		}
	}
	// cout << " a: "; print_vec(a);
	// cout << " num2: " << num_2(a[0]) << endl;
	if(N == 1 ) a.push_back(1);
	ll lcm = LCM(a[0], a[1]);
	for(int i=0; i<N; i++){
		lcm = LCM(lcm, a[i]);
	}
	// cout << " gcd: " << GCD(a[0], a[1]) << " lcm: " << lcm << endl;
	if(lcm > M){
		cout << 0 << endl;
		return 0;
	}

	ll ans = (M/lcm+2-1) /2; // (M/lcm) / 2の切り上げ;
	cout << ans << endl;


}