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;
typedef long long ll;
#define REP(i,n) for(ll i=0, i##_len=(n); i<i##_len; ++i)
#define REPR(i,n) for(ll i = n;i >= 0;--i)
#define FOR(i,m,n) for(ll i = m, i##_len=(n);i <i##_len; ++i)
#define all(x) (x).begin(),(x).end()
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
const int INF = 1e9;
const ll LLINF = 1e16;
 
ll modinv(ll a, ll m) {
    ll b = m, u = 1, v = 0;
    while (b) {
        ll t = a / b;
        a -= t * b; swap(a, b);
        u -= t * v; swap(u, v);
    }
    u %= m;
    if (u < 0) u += m;
    return u;
}

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

    vector<ll> a(n);
    vector<ll> b(n);
    ll kisuc = 0;
    REP(i, n)
    {
        cin >> a[i];
        b[i] = a[i] / 2;
        if(b[i] % 2== 1){
            kisuc++;
        }
    }

    ll lcm = b[0];
    FOR(i,1,b.size()){
        lcm = lcm * b[i] / gcd(lcm, b[i]);
    }

    ll cnt = 0;
    ll kisu = 1;
    while (kisu * lcm <= m)
    {
        cnt++;
        kisu += 2;
    }

    cout << cnt << endl;

    return 0;
}