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;

#define REP(i,m,n) for(int i=(m); i<(int)(n); i++)
#define RREP(i,m,n) for(int i=(int)(n-1); i>=m; i--)
#define rep(i,n) REP(i,0,n)
#define rrep(i,n) RREP(i,0,n)
#define all(a) (a).begin(),(a).end()
#define rall(a) (a).rbegin(),(a).rend()
#define fi first
#define se second
#define debug(...) {cerr<<"[L"<<__LINE__<<"] "; _debug(__VA_ARGS__);}

template<typename T>
string join(const vector<T>&v, string del=", "){ stringstream s;
    for(auto x : v) s << del << x; return s.str().substr(del.size());
}
template<typename T>
ostream& operator<<(ostream& o, const vector<T>&v){
    if(v.size()) o << "[" << join(v) << "]"; return o;
}
template<typename T>
ostream& operator<<(ostream& o, const vector<vector<T> >&vv){
    int l = vv.size();
    if(l){ o<<endl; rep(i,l) o << (i==0 ? "[ " : ",\n  " ) << vv[i] << (i==l-1 ? " ]" : ""); }
    return o;
}
template<typename T1, typename T2>
ostream& operator<<(ostream& o, const pair<T1, T2>& p){
    return o << "(" << p.first << ", " << p.second << ")";
}
inline void _debug(){cerr<<endl;}
template<class First, class... Rest>
void _debug(const First& first, const Rest&... rest){cerr<<first<<" ";_debug(rest...);}

typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<ll> vl;
typedef vector<vl> vvl;

const double PI = (1*acos(0.0));
const double EPS = 1e-9;
const int INF = 0x3f3f3f3f;
const ll INFL = 0x3f3f3f3f3f3f3f3fLL;
const ll mod = 1e9 + 7;

inline void finput(string filename) {
    freopen(filename.c_str(), "r", stdin);
}

int main(){
    ios_base::sync_with_stdio(0);
    // finput("./input");

    auto count_two = [&](ll x){
        int cnt = 0;
        while(x % 2 == 0){
            x /= 2;
            cnt++;
        }
        return cnt;
    };
    auto lcm = [&](ll a, ll b){
        return a*b / __gcd(a, b);
    };
    ll n,m; cin >> n >> m;
    vl a(n), b(n);
    rep(i,n) cin >> a[i];
    rep(i,n) a[i] = a[i] / 2;
    rep(i,n) b[i] = count_two(a[i]);

    if(all_of(all(b), [b](ll x){ return x == b[0]; })){
        ll x = accumulate(all(a), 1, lcm);
        ll i = 1;
        while(x*i <= m) i += 2;
        cout << i / 2 << endl;
    }else{
        cout << 0 << endl;
    }
    return 0;
}