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: Accepted)

#include<iostream>
#include<string>
#include<vector>
#include<algorithm>
#include<bitset>
#include<set>
#include<map>
#include<stack>
#include<queue>
#include<deque>
#include<list>
#include<iomanip>
#include<cmath>
#include<cstring>
#include<functional>
#include<cstdio>
#include<cstdlib>
using namespace std;

#define repr(i, a, b) for (int i = (int)(a); i < (int)(b); i++)
#define rep(i, n) repr(i, 0, n)
#define INF 2e9
#define MOD 1000000007
//#define MOD 998244353
#define LINF (long long)4e18
#define jck 3.141592

const double EPS = 1e-10;

using ll = long long;
using Pi = pair<int,int>;
using Pl = pair<ll,ll>;

ll GCD(ll a, ll b) {
    if (b == 0) return a;
    else return GCD(b, a % b);
}


ll LCM(ll a, ll b){
    return a/GCD(a,b) * b;
}

int main(){
    int N,M; cin >> N >> M;
    vector<int> a(N);
    rep(i,N) cin >> a[i];
    rep(i,N) a[i] /= 2;
    bool ok = true;
    vector<int> b = a;
    vector<int> cnt(N);
    rep(i,N){
        int tmp = 0;
        while(b[i] % 2 == 0){
            b[i] /= 2;
            tmp++;
        }
        cnt[i] = tmp;
    }
    repr(i,1,N){
        if(cnt[i] != cnt[i-1]) ok = false;
    }
    if(!ok){
        cout << 0 << endl;
        return 0;
    }
    ll now = 1;
    rep(i,N){
        now = LCM(now,a[i]);
        if(now > M){
            cout << 0 << endl;
            return 0;
        }
    }
    cout << (ll)(M+now)/(now*2) << endl;

}