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: Runtime Error)

#include <vector>
#include <iostream>
#include <string>
#include <math.h>
#include <algorithm>
#include <map>
using namespace std;

int main() {
  int n,m;
  std::cin >> n>>m;
  std::vector<long long> a(n);
  for (long long i = 0; i < n; i++) {
    long long temp;
    std::cin >> temp;
    a[i]=temp;
  }

  sort(a.begin(), a.end(), std::greater<int>() );//降順
  a.erase(std::unique(a.begin(), a.end()), a.end());

  std::vector<long long> ans;

  for (long long p = 0; m >= a[0]*p+a[0]/2 ; p++) {
    ans.push_back(a[0]*p+a[0]/2);
    // std::cout <<"push:"<<ans[p] <<" p:"<<p<< '\n';
  }

  bool flag=false;
  long long k=0;
  long long now=0;
  if(a.size()>1)
    for (long long i = 1; i < a.size(); i++) {

      flag=false;
      k=0;
      int max= ans[ans.size()-1];
      // std::cout <<'\n'<< max << '\n';

      for (long long p = 0; max >= a[i]*p+a[i]/2 ; p++) {
        // std::cout <<"k:"<<k<< " p:" <<p<< " now:"<<a[i]*p+a[i]/2<<'\n';
        // std::cout <<"ans[k]:"<<ans[k] << '\n';
        now=a[i]*p+a[i]/2;
        if(now==ans[k])
        {
          // std::cout << "Break:" << ans[k] <<'\n';
          k++;
        }
        else if(now>ans[k])
        {
           // cout << "Erase:" <<ans[k]<< '\n';
           ans.erase(ans.begin() + k);
        }
      }

      // std::cout << now<<":"<<ans[ans.size()-1] << '\n';
      while (now<ans[ans.size()-1]) {
        // std::cout << "超過:" <<ans[ans.size()-1]<< '\n';
        ans.erase(ans.begin() + (ans.size()-1));
        // for (int l = 0; l < ans.size(); l++) {
        //   std::cout << " :"<<ans[l] << '\n';
        // }
        // std::cout  << '\n';
      }

    }

  // for (int i = 0; i <ans.size(); i++) {
  //   std::cout <<"ans:"<<ans[i] << '\n';
  // }

  std::cout <<ans.size() << '\n';




  return 0;
}