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;
#define rep(i,n) for(int (i)=0;(i)<(n);(i)++)
#define repp(i,m,n) for(int (i)=(m);(i)<(n);(i)++)
#define repm(i,n) for(int (i)=(n-1);(i)>=0;(i)--)
#define INF (1ll<<60)
#define all(x) (x).begin(),(x).end()
typedef long long lint;
const lint MOD =1000000007;
const lint MAX = 10000000;
using Graph =vector<vector<lint>>;

 
lint fac[MAX], finv[MAX], inv[MAX];

void COMinit() // 二項係数を求める時は前処理としてCOMinit()をする
{
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (lint i = 2; i < MAX; i++)
    {
        fac[i] = fac[i - 1] * i % MOD;
        inv[i] = MOD - inv[MOD % i] * (MOD / i) % MOD;
        finv[i] = finv[i - 1] * inv[i] % MOD;
    }
}
 
// 二項係数計算
long long COM(lint n, lint k)
{
    if (n < k)
        return 0;
    if (n < 0 || k < 0)
        return 0;
    return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}

lint primary(lint num)
{
    if (num < 2) return 0;
    else if (num == 2) return 1;
    else if (num % 2 == 0) return 0; // 偶数はあらかじめ除く

    double sqrtNum = sqrt(num);
    for (int i = 3; i <= sqrtNum; i += 2)
    {
        if (num % i == 0)
        {
            // 素数ではない
            return 0;
        }
    }

    // 素数である
    return 1;
}
  lint lcm(lint a,lint b){
      return a/__gcd(a,b)*b;
  }

  int main(){
      lint N,M;
      cin>>N>>M;
      lint a[N];
      lint b[N];
      rep(i,N){
          cin>>a[i];
          b[i]=a[i]/2;
      }
      bool m=true;
      bool n=true;
      if(N==1){
          lint f=(M-N/2)/N;
          cout<<f;
          return 0;
      }
      lint x=lcm(a[0],a[1]);
      lint y=lcm(b[0],b[1]);
      rep(i,N-1){
            x=lcm(x,a[i+1]);
            if(x>M)m=false;
            y=lcm(y,b[i+1]);
            if(y>M){
                cout<<0;
                return 0;
            }
             
      }
      lint h=0;
      if(y<=M)h++;
      if(M<=y){
          cout<<0;
          return 0;
      }
      lint ans=(M-y)/x+h;
     if(m) cout<<ans;
     else cout<<h;
     
  }