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 <algorithm>
#include <bits/stdc++.h>
using namespace std;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define all(x) (x).begin(), (x).end()
#define ll long long
#define INF 1000000000000000000
typedef pair<ll, ll> pll;

// GCD
/* a と b の最大公約数を返す関数 */
long long GCD(long long a, long long b) {
    if (b == 0)
        return a;
    else
        return GCD(b, a % b);
}

// LCM
long long LCM(long long a, long long b) { return a / GCD(a, b) * b; }

int main() {
    ll N, M;
    cin >> N >> M;
    vector<int> A(N);
    map<ll, ll> ma;
    rep(i, N) {
        cin >> A[i];
        int cnt = 0;
        while (A[i] % 2 == 0) {
            A[i] /= 2;
            cnt++;
        }
        ma[cnt]++;
    }

    if (ma.size() != 1) {
        cout << 0 << endl;
        return 0;
    }

    ll lcm = A[0];
    rep(i, N) { lcm = LCM(lcm, A[i]); }

    if (lcm == 1) {
        cout << 0 << endl;
    } else {
        ll num = M / lcm;
        ll ans = num / 2 + num % 2;
        cout << ans << endl;
    }
}