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>
#include<string>
#include<stdio.h>
#include<deque>
#include<map>
#include<queue>

typedef long long ll;

#define debug(x) cout << #x << "=" << x << endl;
#define debug_arr(a, n) for(ll i = 0; i < n; i++)cout << a[i] << " "
#define inf 100000000000
#define loop(i,n) for(ll i = 0; i < n; i++)
#define graph vector<vector<ll>>
#define P pair<ll,ll>

using namespace std;

ll gcd(ll a,ll b){
    return b? gcd(b,a%b):a;
}

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

int f(ll n){
    ll res = 0;

    while(n % 2 == 0){
        n /= 2;
        res ++;
    }

    return res;
}

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

    vector<ll> a(n);
    loop(i,n) cin >>a[i];

    loop(i,n) {
        if (a[i] % 2 != 0) {
            cout << 0 << endl;
            return 0;
        }
        a[i] /= 2;
    }

    int t = f(a[0]);
    for(int i=1; i < n; i++){
        if(t != f(a[i])){
            cout << 0 << endl;
            return 0;
        }

        a[i] >>= t;
    }
    m >>= t;

    ll l = 1;
    loop(i,n){
        l = lcm(l,a[i]);
        if(l > m){
            cout << 0 << endl;
            return 0;
        }
    }
    m /= l;
    ll ans = (m+1)/2;
    cout << ans << endl;

    return 0;
}