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 <bits/stdc++.h>

using namespace std;
typedef long long LL;
typedef pair<int,int> pii;

#define pb push_back
#define eb emplace_back
#define mp make_pair
#define fi first
#define se second
#define debug(x)  do{cerr<<#x<<": "<<x<<"\n";}while(0)
#define debugv(x) do{cerr<<#x<<": ";for(auto&e: (x))cerr<<e<<" ";cerr<<"\n";}while(0)
#define fori(i, n) for (int i = 0; i < (int)(n); ++i)
#define forn(i, n) for (int i = 1; i <= (int)(n); ++i)
#define ford(i, n) for (int i = (int)(n); i >= 0; --i)
#define fore(i, a, b) for (int i = (int)(a); i <= (int)(b); ++i)
template<typename T> bool mmax(T &m, const T q) { if (m < q) {m = q; return true;} else return false; }
template<typename T> bool mmin(T &m, const T q) { if (m > q) {m = q; return true;} else return false; }
#define RETURN(x) do{cout << x << '\n'; return 0;}while(0)

const int MOD = 1e9+7;
vector<int> primes;
vector<int> smallest;

LL gcd(LL a, LL b) {
    return (b == 0) ? a : gcd(b, a%b);
}

LL lcm(LL a, LL b) {
    LL g = gcd(a,b);
    return a*b/g;
}

int main(int argc, char **argv) {
#ifdef OJ
    freopen("input.txt", "rt", stdin);
    //freopen("output.txt", "wt", stdout);
#endif
    ios::sync_with_stdio(false);
    cin.tie(0);

    int n,m;
    cin >> n >> m;
    vector<int> a(n);
    LL l = 1;
    int odd = 0;
    fori(i,n) {
        cin >> a[i];
        a[i] /= 2;
        if(a[i]%2) odd = 1;
        l = lcm(l,a[i]);
        if(l > m) RETURN(0);
    }

    int ans = m/l;
    if(odd) ans = (ans+1)/2;
    cout << ans << endl;
    return 0;
}