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;
typedef long long ll;

ll gcd(ll a, ll b){return b ? gcd(b,a%b):a;}
ll lcm(ll a, ll b){return a/gcd(a,b)*b;}

int f(int &a){
  int cnt=0;
  while(a%2 == 0){
    a /= 2;
    ++cnt;
  }
  return cnt;
}

int main(){
  int n, m;
  cin >> n >> m;
  vector<int> vec(n);
  for(int i=0; i<n; ++i){
    cin >> vec.at(i);
    if(vec.at(i)%2 != 0){
      cout << 0 << endl;
      return 0;
    }
    vec.at(i) /=2;
  }
  int t=f(vec.at(0));
  for(int i=1; i<n; ++i){
    if(t != f(vec.at(i))){
      cout << 0 << endl;
      return 0;
    }
  }
  for(int i=0; i<t; ++i){
    m /= 2;
  }
  ll l=1;
  for(int i=0; i<n; ++i){
    l = lcm(l, vec.at(i));
    if(l > m){
      cout << 0 << endl;
      return 0;
    }
  }
  int ans=(m+l-1)/(2*l);
  cout << ans << endl;
}