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: Time Limit Exceeded)

#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <cmath>
#include <string>
#include <vector>
#include <iomanip>
#include <queue>
#include <deque>
#include <unordered_map>
#define rep(i,n) for(int i=0;i<n;i++)
#define repn(i,n) for(int i=1;i<=n;i++)
using namespace std;
typedef long long ll;

template<class T>T gcd(T a, T b){if(a<b){a^=b;b^=a;a^=b;}return b?gcd(b,a%b):a;}
template<class T>T lcm(T a, T b){return a/gcd(a,b)*b;}

ll N,M;
ll a[100000];

int main(){
    cin>>N>>M;
    rep(i,N){
        cin>>a[i];
        a[i]/=2;
    }

    ll two=1;
    ll x=a[0];
    while(x%2==0){
        two*=2;
        x/=2;
    }
    repn(i,N-1){
        ll x=a[i];
        if(x%two!=0){
            cout<<0<<endl;
            return 0;
        }
        x/=two;
        if(x%2==0){
            cout<<0<<endl;
            return 0;
        }
    }

    ll l=a[0];
    repn(i,N-1){
        l=lcm(l,a[i]);
    }

    int ans=M/l;
    ans=ans&1?ans/2+1:ans/2;
    cout<<ans<<endl;
}