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: Wrong Answer)

#include <bits/stdc++.h>
#define rep(i, n) for(int i = 0; i < (int)(n); ++i)
using namespace std;
using ll = long long;

int gcd(int a, int b) {
    if(b == 0) return a;
    else return gcd(b, a%b);
}

int main() {

    int n, m;
    cin >> n >> m;

    vector<int> a(n);
    vector<int> ac(n);
    rep(i, n) {
        int c;
        cin >> c;
        c /= 2;
        a[i] = c;
        ac[i] = c;
    }

    vector<int> divide2(n);
    rep(i, n) {
        int ex_2 = 0;
        while(ac[i] % 2 == 0) {
            ac[i] /= 2;
            ++ex_2;
        }
        divide2[i] = ex_2;
    }

    ll cc = ll(divide2[0]) * n;
    ll t = 0;


    rep(i, divide2.size()) t += divide2[i];
    if(cc != t) {
        cout << 0 << endl;
        return 0;
    }
    else {
        ll lcm = 1;
        for(int i = 0; i < (int)a.size(); ++i) {
            ll g = gcd(lcm, a[i]);
            lcm = lcm / g * a[i];
        }
        
        ll cnt = 0;
        for(int i = 1; i * lcm <= m; i += 2) {
            ++cnt;
        }
        cout << cnt << endl;
        return 0;
    }
}