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 <bits/stdc++.h>
using namespace std;
typedef long long  ll;
#define FOR(i,a,n) for(ll i=(ll)a;i<(ll)n;i++)
#define RFOR(i,a,n) for(ll i=(ll)n-1;i >= (ll)a;i--)
#define rep(i,n) FOR(i,0,n)
#define rrep(i,n) RFOR(i,0,n)
#define ALL(v) v.begin(), v.end()
ll MOD = 1000000007;
ll INF = 214748364700000000;
double EPS = 1e-12;
double PI = 3.141592653589793238;
template<typename T>
void remove(std::vector<T>& vector, unsigned int index){
    vector.erase(vector.begin() + index);
}

using Graph = vector<vector<ll>>;

ll GCD(ll a,ll b){
    if(a < b) return GCD(b,a);
    ll r = INF;
    while(r){
        r = a % b;
        a = b;
        b = r;
    }
    return a;
}

int main(){
   ll N,M;cin >> N >> M;
   vector<ll> A(N);
   ll m = 0;
   rep(i,N){
       cin >> A[i];
   }
   sort(ALL(A));
   vector<ll> B{A[0]};
   FOR(i,1,N){
       if(A[i-1] != A[i]) B.emplace_back(A[i]);
   }
   N = B.size();
   ll gcd = A[0],lcm = 1;
   rep(i,N) gcd = GCD(gcd,B[i]);
   rep(i,N){
       lcm *= B[i] / gcd;
   }
   lcm *= gcd;
   ll a = 2*M / lcm;
   cout << (a+1) / 2 << endl;
}