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 9.2.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
using P = pair<int,int>;
/* ちゃんと考えてわかって実装 */

// 最大公約数
ll gcd(ll a, ll b){
    if(b == 0) return a;
    return gcd(b, a % b);
}

// 最小公倍数(overflowに注意した)
ll lcm(ll a, ll b){
    return a / gcd(a, b) * b;
}

ll cnt2(ll a){
    ll cnt = 0;
    while(a % 2 == 0){
        a /= 2;
        cnt++;
    }
    return cnt;
}

int main(void){
    ll n, m;
    cin >> n >> m;
    vector<ll> a(n);
    ll Xmin = 1;
    ll num = 1;
    for(ll i=0; i<n; i++){
        cin >> a[i];
        // Xmin = lcm(Xmin, a[i]/2);
        if(i!=0 && cnt2(a[i]) == cnt2(a[i-1])){
            num++;
            a[i] /= cnt2(a[i]);
            Xmin = lcm(Xmin, a[i]);
        }
    }

    ll ans;
    m /= cnt2(a[0]);
    if(m - Xmin < 0 || num != n) ans = 0;
    else{
        ans = 1LL + floor(double(m - Xmin) / double(2*Xmin));
    }
    cout << ans << endl;
    return 0;
}