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)

/*
これを入れて実行
g++ code.cpp
./a.out
 */
 
#include <iostream>
#include <stdio.h>
#include <vector>
#include <string>
#include <queue>
#include <deque>
#include <algorithm>
#include <utility>
#include <set>
#include <map>
#include <cmath>
#include <math.h>
#include <tuple>
#include <iomanip>
#include <bitset>
 
using namespace std;
typedef long long ll;
typedef long double ld;
 
int dy4[4] = {-1, 0, +1, 0};
int dx4[4] = {0, +1, 0, -1};
int dy8[8] = {-1, -1, 0, 1, 1, 1, 0, -1};
int dx8[8] = {0, 1, 1, 1, 0, -1, -1, -1};
 
const long long INF = 1LL << 60;
const ll MOD = 1e9 + 7;
 
bool greaterSecond(const pair<int, int>& f, const pair<int, int>& s){
    return f.second > s.second;
}
 
ll gcd(ll a, ll b){
	if (b == 0)return a;
	return gcd(b, a % b);
}
 
ll lcm(ll a, ll b){
    return a / gcd(a, b) * b;
}
 
ll nCr(ll n, ll r){
    if(r == 0 || r == n){
        return 1;
    } else if(r == 1){
        return n;
    }
    return (nCr(n - 1, r) + nCr(n - 1, r - 1));
}
 
ll nPr(ll n, ll r){
    r = n - r;
    ll ret = 1;
    for (ll i = n; i >= r + 1; i--) ret *= i;
    return ret;
}
 
//-----------------------ここから-----------

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

    vector<ll> a(n);
    for(int i = 0; i < n; i++) cin >> a[i];

    bool f = true;
    ll maxi = 0;
    ll lcmAll = 1;

    set<ll> div;


    for(int i = 0; i < n; i++){
        maxi = max(maxi, a[i]);
        lcmAll = lcm(lcmAll, a[i] / 2);
        if((a[i] / 2) % 2 == 0){
            ll cnt = 0;
            ll tmp = a[i];
            while(true){
                if(tmp % 2 == 0){
                    tmp /= 2;
                    cnt++;
                } else {
                    break;
                }
            }
            div.insert(cnt);
        } else {
            div.insert(1);
        }
    }

    if(div.size() > 1){
        f = false;
    }

    if(!f){
        cout << 0 << endl;
        return 0;
    }

    ll ans = 0;


    ans = (m / lcmAll + 1) / 2;


    cout << ans << endl;


}