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: Accepted)

#include <bits/stdc++.h>
using namespace std;

int n, m;
vector<int>a;
vector<int>b;

void input(){
	
	scanf("%d %d", &n, &m);
	
	a.resize(n);
	b.resize(n);
	
	for(int ai, i = 0 ; i < n ; i ++){
		scanf("%d", &ai);
		a[i] = ai;
	}
	
	return;
}

int solve(){
	
	int q2 = -1;
	for(int i = 0 ; i < n ; i ++){
		int ai = a[i];
		if(q2 == -1){
			q2 = 0;
			while((ai % 2) == 0){
				ai /= 2;
				q2 ++;
			}
		} else {
			int curr_q2 = 0;
			while((ai % 2) == 0){
				ai /= 2;
				curr_q2 ++;
			}
			if(q2 != curr_q2) return 0;
		}
		b[i] = ai;
	}
	
	long long X_1 = (1<<(q2 - 1));
	long long lcm = 0LL;
	
	for(int i = 0 ; i < n ; i ++){
		int bi = b[i];
		if(!lcm) lcm = bi;
		else lcm = (lcm*bi)/__gcd(lcm, (long long)bi);
		if(lcm > m) return 0;
	}
	
	X_1 *= lcm;
	long long cnt = 0LL;
	
	for(long long i = 1 ; ; i += 2){
		long long X = i*X_1;
		if(X <= m) cnt++;
		else break;
	}
	
	return cnt;
}

int main(){
	
	input();
	printf("%d\n", solve());
	
	return 0;
}