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: Accepted)

#include <bits/stdc++.h>
typedef long long ll;
using namespace std;
ll MOD = 1000000007;

#define vec              vector<int>
#define vecll            vector<ll>
#define vecd             vector<double>
#define vecst            vector<string>
#define vecb             vector<bool>
#define v2(v,n,m,init)   vector<vector<int>>  v(n, vector<int>(m, init))
#define vb2(v,n,m,init)  vector<vector<bool>> v(n, vector<bool>(m, init))
#define vll2(v,n,m,init) vector<vector<ll>>   v(n, vector<ll>(m, init))

#define rep(i,n)         for(ll i=(ll)0; i<(ll)n; i++)
#define REP(i,m,n)       for(ll i=(ll)m; i<(ll)n; i++)
#define arr(var, n)      vec var(n); rep(i,n){cin >> var[i];}
#define arrll(var, n)    vecll var(n); rep(i,n){cin >> var[i];}
#define arrst(var, n)    vecst var(n); rep(i,n){cin >> var[i];}
#define all(var)         (var).begin(), (var).end()
#define sortall(var)     sort(all(var))
#define uniqueall(v)     v.erase(unique(v.begin(), v.end()), v.end());
#define pb(var)          push_back(var)
#define prt(var)         cout << (var) << "\n"
#define prtd(n, var)     cout << fixed << setprecision(n) << (var) << "\n"
#define prtfill(n, var)  cout << setw(n) << setfill('0') << (var);
#define prt2(v1, v2)     cout << (v1) << " " << (v2) << "\n"
#define prt3(v1, v2, v3) cout << (v1) << " " << (v2) << " " << (v3) << "\n"
#define prtall(v)        rep(i,v.size()){cout<<v[i]<<(i!=v.size()-1?" ":"\n");}
template <typename T>
bool chmax(T &a, const T& b){if(a<b){a=b; return true;} return false;}
template <typename T>
bool chmin(T &a, const T& b){if(a>b){a=b; return true;} return false;}
//------------------------------------------------------------------
ll gcd(ll a, ll b){ return b?gcd(b,a%b):a;}
ll lcm(ll a, ll b){ return a/gcd(a,b)*b;}

ll div2(ll x){
	ll num=0;
	while(x%2==0){
		num++;
		x/=2;
	}
	return num;
}

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

	rep(i,n){
		a[i]/=2;
	}

	ll k=div2(a[0]);
	rep(i,n){
		if(k!=div2(a[i])){
			prt(0);
			return 0;
		}
		a[i]>>=k;
	}
	m>>=k;


	ll l=1;
	rep(i,n){
		l = lcm(a[i], l);
		if(l>m){
			prt(0);
			return 0;
		}
	}

	prt((m-l)/(2*l)+1);

}