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 (int i=0; i < (int)(n); i++)
using ull = unsigned long long;
using ll = long long;

ll gcd(ll a, ll b) {
    if (a%b==0) return b;
    else return gcd(b, a%b);
}
ll lcm(ll a, ll b) {
    return a * b / gcd(a, b);
}

int main() {
    ll N, M;
    cin >> N >> M;

    ll a[N];
    rep(i, N) {
        cin >> a[i];
        a[i] = a[i] / 2;
    }

    ll minX = 1;
    bool over = false;
    rep(i, N) {
        if (minX > M) {
            over = true;
            break;
        }
        minX = lcm(minX, a[i]);
    }

    if (over) cout << 0 << endl;
    else cout << M / minX - M / (2 * minX) << endl;

    return 0;
}