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 pb push_back
#define eb emplace_back
#define fi first
#define se second
#define endl '\n'
#define fcin ios::sync_with_stdio(false); cin.tie(nullptr);
#define INF 0x3f3f3f3f
#define all(x) x.begin(), x.end()
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

typedef unsigned long long ll;
typedef pair<int, int> pii;

ll gcd(ll a, ll b){
    return b == 0 ? a : gcd(b, a%b);
}

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

int main(){
    fcin;
    ll n, m;
    cin >> n >> m;

    vector<ll> a(n);
    for(ll i = 0; i < n; i++) cin >> a[i];

    ll L = a[0]/2;
    for(ll i = 1; i < n; i++) L = lcm(L, a[i]/2);
    
    for(ll i = 0; i < n; i++) if((L / (a[i] / 2)) % 2 == 0) return cout << 0, 0;

    ll l = 1, r = m;
    ll ans = 0;
    while(l <= r){
        ll mid = (l + r) / 2;

        if(L + 2LL * mid * L <= m){
            ans = max(ans, mid);
            l = mid + 1;
        } else r = mid - 1;
    }

    cout << ans + (r == m) + 1 << endl;
    return 0;
}