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>
#define INF 1e9
#define llINF 1e18
#define MOD 1000000007
#define pb push_back
#define mp make_pair 
#define F first
#define S second
#define ll long long
#define ull unsigned long long
#define vi vector<ll>
#define vvi vector<vi>
#define BITLE(n) (1LL<<((ll)n))
#define BITCNT(n) (__builtin_popcountll(n))
#define SUBS(s,f,t) ((s).substr((f)-1,(t)-(f)+1))
#define ALL(a) (a).begin(),(a).end()
using namespace std;
ll gcd(ll a,ll b){
  ll r=a%b;
  while(r>0){
    a=b;b=r;r=a%b;
  }
  return b;
}
ll lcm(ll a,ll b){
  return a/gcd(a,b)*b;
}
int main(){
  cin.tie(0);
  ios::sync_with_stdio(false);
  ll n,m;cin>>n>>m;
  ll now;
  cin>>now;
  bool fl=false,fl2=false;
  if(now%4==0)fl2=true;
  else fl=true;
  bool anfl=false;
  for(int i=1;i<n;i++){
    ll a;cin>>a;
    if(a%4==0)fl2=true;
    else fl=true;
    now=lcm(now,a);
    if(now/2 > m){anfl=true;}
  }

  if(anfl){
    cout<<0<<endl;return 0;
  }
  if(fl&&fl2){
    cout<<0<<endl;return 0;
  }
  ll hoge=now/2;
  if(hoge>m)cout<<0<<endl;
  else{
    cout<<1+(m-hoge)/now<<endl;
  }
  return 0;
}