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 <cstring>
#include<string>
#include<algorithm>
#include<vector>
#include<iomanip>
#include<math.h>
#include<complex>
#include<queue>
#include<deque>
#include<stack>
#include<map>
#include<set>
#include<bitset>
#include<functional>
#include<assert.h>
#include<numeric>
using namespace std;
#define REP(i,m,n) for(int i=(int)(m) ; i < (int) (n) ; ++i )
#define rep(i,n) REP(i,0,n)
using ll = long long;
const int inf=1e9+7;
const ll longinf=(1LL<<63) - 600;
const ll mod=1e9+7 ;

ll gcd(ll x, ll y){
  if(x%y==0) return y;
  else return gcd(y, x%y);
}

int main(){
  int N; ll M; cin >> N >> M; ll a[N];
  rep(i, N){
    cin >> a[i]; 
    if (a[i]%2 == 1) {
      cout << 0 << endl;
      return 0;
    }
    a[i]/=2;
  }
  vector<int> power(N); ll tmp = 1;
  rep(i, N){
    int X = a[i];
    while(!(X%2)){
      X/=2; power[i]++;
    }
    if(i>=1&&power[i-1]!=power[i]){
      cout << 0 << endl;
      return 0;
    }
    tmp = tmp*a[i]/gcd(tmp, a[i]);
  }
  if(tmp>M) cout << 0 << endl;
  else cout << M/(2*tmp)+1 << endl;  
}