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>
#pragma GCC optimize ("Ofast")
#pragma GCC target ("avx2")
#pragma GCC optimization ("O3")
#pragma GCC optimization ("unroll-loops")
using namespace std;
#define F first
#define S second
#define int long long

///My functions
int lcm (int a, int b) { return ( a / __gcd ( a , b ) * b ) ; }

//Constants
const int mxn = 3*1e5+5, inf = 1e9+7, mod = 1e9+7;
const long long infLL = 1e18+5;
const double EPS = 1e-9 , PI = acos(-1);

//Structs

//Variables
int n,m,Q,k,a,b,c,sum,ans,MAXans,MINans,sc,pos,acc[mxn],racc[mxn],ar[mxn];
vector<int>v,g[mxn];
bitset<mxn>flag;
bool sol ;
string s;

int32_t main(){
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    
    cin >> n >> m ; m *= 10 ; sc = 5 ;
    for( int i = 1 ; i <= n ; i++ ) {
    	cin >> ar[i] ; 
    	ar[i] *= 10 ; 
    	sc = lcm( ar[i] , sc ) ;
    }

    //cout << sc << ' ' ; 

    cout << ( m / sc ? m / sc + 1 : 0 ) ;





    return 0;
}