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 <string>
#include <algorithm>
#include <vector>
#include <map>
#include <math.h>

typedef long long ll;
using namespace std;
 
 
ll GreatestCommonDivisor(ll p, ll q)
{
  if ( p < q ) swap( p, q);
  while ( true )
  {
    long long yo = p % q;
    if ( yo == 0 ) break;
    p = q;
    q = yo;
  }
  return q;
}
 
//最小公倍数
long long LeastCommonMultiple(ll a, ll b)
{
  return a / GreatestCommonDivisor(a,b) * b;
}
 
 
ll f(ll a)
{
  ll num = 0;
  while ( a > 0 )
  {
    if ( a % 2 == 1) return num;
    ++num;
    a = a / 2;
  }
  return num;
}
 

int main()
{
  ll n, m;
  cin >> n >> m;
  
  vector<ll> a(n);
  for ( auto& ai : a) {
    cin >> ai;
    ai /= 2;
  }
 
 
  //ai に 含まれる2の数を計算（同じでないと半公倍数は存在せず）
  ll c = f( a[0] );
  for ( int i = 1; i < n; ++i )
  {
    if ( c != f(a[i])) { 
      cout << 0;
      return 0;
    }
  }
 
  //a の最小公倍数を計算
  ll lcm = a[0];
  for ( ll i = 1; i < n; ++i )
    lcm = LeastCommonMultiple(lcm, a[i]);

  if ( lcm > m ) {
    cout << 0;
    return 0;
  }

  cout << (m -lcm) / 2 / lcm + 1;  
  return 0;
}