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;
//#include "testlib.h"
#define ff first
#define ss second
#define all(v) v.begin(),v.end()
#define int long long
#define ll long long
#define M 1000000007
#define MM 998244353
#define inputarr(a,n) for(int i=0;i<n;++i) cin>>a[i]
#define GCD(m,n) __gcd(m,n)
#define LCM(m,n) m*(n/GCD(m,n))
#define mii  map<ll ,ll >
#define sz(a) (int)a.size()
#define msi  map<string,ll >
#define rep(a,b)    for(ll i=a;i<b;i++)
#define rep0(n)    for(ll i=0;i<n;i++)
#define repi(i,a,b) for(ll i=a;i<b;i++)
#define pb push_back
#define vi vector<ll>
#define vs vector<string>
#define ppb pop_back
#define endl '\n'
#define asdf ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define r0 return 0;
#define FORD(i, a, b) for (int i = (int) (a); i >= (int) (b); --i)
#define inputoutput freopen("input.txt", "r", stdin);freopen("output.txt", "w", stdout);
#define Set(a, s) (a, s, sizeof (a))
#define FOR repi
#define pii pair<int,int>
#define REVERSE(v) reverse(all(v))
#define trav(a, x) for(auto& a : x)
#define display(x) trav(a,x) cout<<a<<" ";cout<<endl
#define debug cerr<<"bhau"<<endl
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
    std::cerr << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
    const char* comma = strchr(names + 1, ',');std::cerr.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}
template<typename T, typename U> static inline void amin(T &x, U y) 
{ 
    if (y < x) 
        x = y; 
}
template<typename T, typename U> static inline void amax(T &x, U y) 
{ 
    if (x < y) 
        x = y; 
}

ll max(ll a, ll b) { return (a > b)? a : b;}
int min(int a, int b) { return (a < b)? a : b;} 


    
     int solve()
     {  
         int n,k;cin>>n>>k;
        // set<int> s;
         int l;cin>>l;
         l/=2;
         int t2=l;
        rep(1,n){
            int x;cin>>x;
            if(x%2) return cout<<0,0;
            // s.insert(x);
            l=(l*x/2)/__gcd(l,x/2);
        }
        trace(l);
        int p=k/l;
        int num=(k-1)/(2*l);
        num+=1;
        cout<<num<<endl;

        r0   
     }
    signed main()
    {   
        asdf
        
        int t=1;
        //cin>>t;
        while(t--)
        {
            solve();
        }     
        
}