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 _GLIBCXX_DEBUG
#define pb push_back;
#define all(x) (x).begin(), (x).end()
typedef long long ll;
typedef pair<char, char> c_c;
typedef pair<int, int> i_i;
typedef pair<ll, ll> l_l;
const int INF = 1e9;
const ll LLINF = 1e16;

int gcd(int x, int y){
	if(x>y) swap(x, y);
	while(1){
		y %= x;
		swap(x, y);
		if(x==0) return y;
	}
}

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    
    int n;
    ll m;
    cin >> n >> m;
    vector<int> a(n);
    for(int i=0; i<n; i++) cin >> a[i];
    set<int> s;
    vector<int> b(n);
    b = a;
    for(int i=0; i<n; i++){
        int cnt = 0;
        while(a[i]%2 == 0){
            a[i] /= 2;
            cnt++;
        }
        s.insert(cnt);
        //b[i] = a[i];
    }
    ll sum = b[0];
    if (s.size() != 1){
        cout << 0 << endl;
        return 0;
    }
    else{
        //cout << LC << endl;
        for(int i=1; i<n; i++){
            sum = (sum*b[i])/(gcd(sum, b[i]));
        }
        sum /= 2;//LCM求め終わり
        //sum *= (*begin(s));
    }
    int ans = 0;
    //cout << sum << endl;
    ll tot = sum;
    while(tot <= m){
        ans++;
        tot += sum*2;
    }
    cout << ans << endl;
}