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<stdio.h>
#include<stdlib.h>
#include<iostream>
#include<string>
#include<algorithm>
#include<vector> 
#include<limits>
#include<numeric>
#include<type_traits>
#include<math.h>
#include<fstream>
#include<set>
#include<queue>
using namespace std;
typedef long long ll;

#define rep(i, n) for(int i = 0; i < (int)(n); i++)
#define repi(i,a,b) for(int i = (int)(a); i < (int)(b); i++)
#define out(str) cout << str << endl
#define ALL(a) (a).begin(),(a).end()
#define INF (1<<29)

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; }

ll gcd(ll a, ll b) {
  return b != 0 ? gcd(b, a % b) : a;
}
ll lcm(ll a, ll b) {
  return a / gcd(a, b) * b;
}

bool is_product_overflow(long a, long b) {

	long prod = a * b;
	
	return (prod / b != a);
}

ll N,M,A=1,tmpA;

int main(){
    cin>>N>>M;
    rep(i,N){
        cin>>tmpA;
        
        A=lcm(A,tmpA);
        if(A>M||is_product_overflow(A,tmpA)){
            out(0);
            return 0;
        }
    }
    out((M+A/2)/A);
    

}