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 rep(i, n) for (int64_t i = 0; i < (int64_t)(n); i++)

int main(){
    int64_t n, m;
    cin >> n >> m;
    vector<int64_t> a(n);
    rep(i,n) cin >> a[i];

    vector<int64_t> b(n);
    bool bl = true;
    int64_t ans = 0;
    rep(i,n){
        int64_t ai = a[i];
        int64_t bi = 0;
        while(ai%2 != 0){
            ai /= 2;
            bi++;
        }
        b[i] = bi;
        if(i > 0 && b[i-1] != b[i]){
            bl = false;
            break;
        }
    }
    int64_t lcma = a[n-1];
    if(bl){
        rep(i,n-1){
            int64_t ai = a[i]; 
            lcma = ai/__gcd(ai,lcma)*lcma;
        }
        lcma /= 2;
        ans = (m/lcma+1)/2;
        if(ans < 0) ans = 0;
    }
    cout << ans << endl;
    //cout << lcma << endl;
}