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: Accepted)

#include<bits/stdc++.h>
#define ll long long
#define double long double
#define rep(i,n) for(ll i=0;i<(n);i++)
#define REP(i,n) for(int i=1;i<=(n);i++)
#define mod (ll)(1e9+7)
#define inf (ll)(3e18+7)
#define pi (double) acos(-1)
#define P pair<int,int>
#define PiP pair<int,pair<int,int>>
#define all(x) x.begin(),x.end()
#define rall(x) x.rbegin(),x.rend()
using namespace std;

int gcd(int a, int b) {
	if (!b)return a;
	return gcd(b, a % b);
}
int lcm(int a, int b) {
	return a / gcd(a, b) * b;
}

int main() {
    ll n, m, l = 1;
    cin >> n >> m;
    vector<ll> a(n);
    rep(i, n){
        cin >> a.at(i);
        l = lcm(l, a.at(i));
    }
    sort(rall(a));
    a.erase(unique(all(a)), a.end());
    ll now = a[0]/2;
    while(m >= now){
        bool ok = true;
        rep(i, n){
            if((now-(a[i]/2))%a[i] != 0)ok = false;
            if(!ok)break;
        }
        if(ok)break;
        now += a[0];
        if(m < now){
            cout << 0 << endl;
            exit(0);
        }
    }
    cout << (m - now + l) / l << endl;
}