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 _GLIBCXX_DEBUG //これつけるとA[N]でもいいらしい
//for文のマクロ
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define big 1000000007
//最大公約数　最小公倍数はa*b/GCD(a,b)
int64_t GCD(int64_t a,int64_t b){
    return b == 0 ? a : GCD(b, a % b);
   
}
int64_t LCM(int64_t a,int64_t b)
{
    return a / GCD(a, b) * b;
}
int main(){
    int64_t n,m;
    cin >> n >> m;
    vector<int64_t> a(n);
    for(int i=0;i<n;i++){
        cin >> a[i];
        a[i] = a[i]/2;
    }
    int64_t lcmm = a[0];
    for(int i=1; i<n; i++){
        lcmm = LCM(lcmm, a[i]);
    }
    //最小公倍数15⇒15,45がそう
    int64_t ans;
    ans = (m/lcmm+1)/2;
    cout << ans << endl;
    
}