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 <vector>

#define ll long long
#define REP(i,n) for(ll i = 0; i < n; i++)
#define PI 3.141592653589

using namespace std;

ll gcd(ll a, ll b){
    if(a > b){
        ll r = a % b;
        while(r != 0){
            a = b;
            b = r;
            r = a % b;
        }
        return b;
    } else{
        ll r = b % a;
        while(r != 0){
            b = a;
            a = r;
            r = b % a;
        }
        return a;
    }
}

ll lcm(ll a, ll b){
    return a * b / gcd(a, b);
}

ll lcm_array(ll array[], ll n){
    if(n >= 2){
        ll x = lcm(array[0], array[1]);
        for(ll i = 2; i < n; i++){
            x = lcm(x, array[i]);
        }
        return x;
    } else return array[0];
}

int main(void){
    
    ll n, m;
    cin >> n >> m;

    ll a[n];
    REP(i,n) cin >> a[i];
    
    ll x = lcm_array(a, n);

    ll ctr = 0;
    for(ll j = 1; ; j += 2){
        if(x * j / 2 > m) break;
        ctr++;
    }

    cout << ctr;
    
    return 0;
}