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 <algorithm>
#include <numeric>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <map>
#include <set>
#include <unordered_map>
#include <unordered_set>
#include <iomanip>
#include <functional>
using namespace std;
#define int long long
#define All(v) (v).begin(),(v).end()
int dy[8] = {-1,0,1,0,-1,1,1,-1};
int dx[8] = {0,1,0,-1,1,1,-1,-1};
const int mod = 1000000007;
const int inf = mod*mod;
const int d5 = 100100;
int gcd(int x, int y) {
    if (y == 0)return x;
    else return gcd(y, x % y);
}
signed main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n,m;
    cin>>n>>m;
    vector<int> a(n),b(n);
    for(int i=0;i<n;i++)cin>>a[i];
    for(int i=0;i<n;i++)b[i]=a[i]/2;
    int g=b[0],l=b[0];
    int G=a[0];
    for(int i=0;i<n;i++){
        l=l*b[i]/gcd(l,b[i]);
    }
    for(int i=0;i<n;i++){
        G=gcd(G,a[i]);
    }
    int x=m%(l*G);
    cout<<m/(l*G)+(x>l)<<endl;
}