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<iostream>
#include<vector>
#include<cmath>
#include<algorithm>
using namespace std;

int gcd(int x,int y){
  if(x<y) swap(x,y);
  //xの方が常に大きい
  while(y>0){
    int r=x%y;
    x=y;
    y=r;
  }
  return x;
}

int lcm(int x,int y){
  return int(x*y/gcd(x,y));
}

//n個の場合のlcmも2個で比べて順に適用していくだけ
int main(){
  int n,m;cin >> n >> m;
  vector<int> a(n);for(int i=0;i<n;i++){cin >> a[i];a[i]/=2;}
  int ans=lcm(a[0],a[1]);//cout << ans << endl;
  for(int i=1;i<n;i++){ans=lcm(ans,a[i]);}
  //cout << ans << endl;
  int k=ans;int l=0;//cout << ans << endl;
  while(k%2==0){
    l+=1;k=k/2;
  }
  //cout << l << endl;
  int f=0;
  for(int i=1;i<n;i++){
    if(a[i]%(int(pow(2,l)))!=0){f=1;break;}
  }
  if(f==1){
    cout << 0 << endl;
  }else{
    cout << int(floor((floor(m/ans)+1)/2)) << endl;
  }
}