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)

//#define _GLIBCXX_DEBUG
#include <bits/stdc++.h>

using namespace std;

//----------------------- Print Function ----------------------//

inline void print() {
    cout << '\n';
}
template <typename First, typename... Rest>
void print(const First &first, const Rest &... rest) {
    cout << first << ' ';
    print(rest...);
}

//------------------------- Libraries -------------------------//

long long gcd(long long a, long long b){
    if (a < b) swap(a, b);
    return b ? gcd(b, a % b) : a;
}

long long lcm(long long a, long long b){
    return a * b / gcd(a, b);
}

long long extGCD(long long a, long long b, long long &x, long long &y) {
    if (b == 0) {
        x = 1;
        y = 0;
        return a;
    }
    long long d = extGCD(b, a%b, y, x);
    y -= a/b * x;
    return d;
}

//--------------------------- Solve ---------------------------//

void solve() {
    long long N, M; cin >> N >> M;
    vector<long long> v(N);
    for (int i = 0; i < N; i++) cin >> v[i];
    sort(v.rbegin(), v.rend());

    long long lcm_ = v[0];
    long long half_lcm = lcm_ / 2;
    for (int i = 1; i < N; i++) {
        long long a = lcm_, b = v[i];
        long long c = b / 2 - half_lcm;
        if (c % gcd(a, b) != 0) {
            cout << 0 << '\n';
            return;
        }
        long long x, y;
        extGCD(a, -b, x, y);
        x *= c / gcd(a, b);
        y *= c / gcd(a, b);
        
        half_lcm = a * x + half_lcm;
        lcm_ = lcm(lcm_, v[i]);

        if (half_lcm > M) {
            cout << 0 << '\n';
            return;
        }
    }

    int ans = (M - half_lcm) / lcm_ + 1;
    
    cout << ans << '\n';
}

int main() {
    cin.tie(nullptr);
    ios_base::sync_with_stdio(false);

    solve();

    return 0;
}