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++ (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <iostream>
#include <cstring>
#include <cmath>
#include <vector>
#include <algorithm>
#include <stdio.h>
#include <math.h>
#include <stdlib.h>
#include <set>
#include <map>
#include <iterator>
#include <stack>
#include <string.h>
#include <cstdlib>
#include <queue>
#include <list>
#include <string>

using namespace std;

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

long long lcm(long long x,long long y)
{
    return x/gcd(x,y) * y;
}

long long rutwo(long long x)
{
    long long r = 0;
    while (x%2 == 0){
        x = x/2;
        r++;
    }
    return r;
}

int main()
{
    long long n,m;
    cin >> n >> m;
    vector<long long>  a;
    long long k;
    for (int i = 0;i < n;i++){
        cin >> k;
        a.push_back(k);
    }
    
    
    sort(a.begin(),a.end(),greater<long long>());
    
    bool check = true;
    
    long long tworui[100005];
    
    for (int i = 0;i < n;i++){
        tworui[i] = rutwo(a[i]);
    }
    
    for (int i = 0;i < n-1;i++){
        if (tworui[i] != tworui[i+1]){
            check = false;
            cout << 0 << endl;
            break;
        }
    }
    
    if (check){
        long long g = a[0]/2;
        for (int i = 0;i < n;i++){
            g = lcm(g,a[i]/2);
        }
    
        cout << (m+g)/(2*g) << endl;
    }
}