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>
// #include <atcoder/all>
#define ll long long int
#define ld long double
#define yorn(f) cout<<((f)?"Yes":"No")<<endl;
#define YORN(f) cout<<((f)?"YES":"NO")<<endl;
#define rep(i, n) for (ll i = 0; i < (ll)(n); i++)
#define repi(i, n) for (ll i = 1; i < (ll)(n); i++)
#define all(x) (x).begin(), (x).end()
#define put(x) cout << x << endl;
#define println(x) cout << x << endl;
using namespace std;
// using namespace atcoder;

long long GCD(long long a, long long b) {
    if (b == 0) return a;
    else return GCD(b, a % b);
}


void Main()
{
    ll n, m;
    cin >> n >> m;

    vector<ll> a(n);
    rep(i, n) cin >> a[i];

  //  ll v = a[0];

    ll v = a[0] / 2;
    repi(i, n) {
        ll gcd = GCD(v, a[i]);
        v = v * (a[i]) / gcd;
    }

    // ll cnt = 0;
    // ll t = a[0];
    // while (t % 2 == 0) {
    //     t /= 2;
    //     cnt++;
    // }
    // ll refCnt = cnt;
    // repi(i, n) {
    //     ll cnt = 0;
    //     ll t = a[i];
    //     while (t % 2 == 0) {
    //         t /= 2;
    //         cnt++;
    //     }
    //     if (cnt != refCnt) {
    //         put(0);
    //         return;
    //     }
    // }

    // put(1 + (m - v) / (2 * v));

    v /= 2;
    rep(i, n) {
        if (v % a[i] != a[i] / 2) {
            put(0);
            return;
        }
    }

   put(1 + (m - v) / (2 * v));


}
signed main(){ Main();return 0;}