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>
using namespace std;
#include <iostream>
#include <vector>
#include <cstdlib>
#include <cmath>
#include <math.h>
#include <sstream>
#include <numeric>
#include <cctype>
#include <bitset>
#include <cassert>
#include<algorithm>
//---------------------------------------------------------------
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define rep1(i, n) for (int i = 1; i <= (int)(n); i++)
#define rep2(i, s, n) for (int i = (s); i < (int)(n); i++)
#define SIZE 100005
#define INF 1000000000000000LL
#define all(x) x.begin(),x.end()
using pint = pair<int,int>;
using vec = vector<int>;
using ll=long long;
using vll = vector<ll>;
using vvll = vector<vector<ll>>;
//---------------------------------------------------------------
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
//---------------------------------------------------------------

//↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);

    ll n,m;cin>>n>>m;
    vll a(n);
    rep(i,n) {cin>>a[i];a[i]/=2;}
    vll b(n,0);
    auto c = a;
    for(int i=0; i<n; i++){
        while(c[i]%2==0){
            c[i]/=2;
            b[i]++;
        }
    }

    bool ok = true;
    rep(i,n-1){
        if(b[i]!=b[i+1]) ok = false;
    }

    if(ok==false){
        cout<<0<<endl;
        return 0;
    }else{
        
        ll t=1;
        rep(i,n){
            t = t*a[i]/__gcd(t,a[i]);
        }

        if(t>m){
            cout<<0<<endl;
            return 0;
        }

        ll ans=1;
        ll cnt = 0;
        while(true){
            cnt++;
            ll T;
            T = t*ans;
            if(T>m){
                cnt--;
                break;
            }
            if(T==m) break;
            ans+=2;
        }

        cout<<cnt<<endl;
        return 0;

    }
}