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)

// Etavioxy
#include<cstdio>
#include<cctype>
#include<cstring>
#include<algorithm>
#include<cmath>
#define il inline
#define ll long long
#define rep(i,s,t) for(register int i=(s);i<=(t);i++)
#define rev_rep(i,s,t) for(register int i=(s);i>=(t);i--)
#define each(i,u) for(int i=head[u];i;i=bow[i].nxt)
#define file(s) freopen(s".in" ,"r",stdin),freopen(s".out","w",stdout)
#define pt(x) putchar(x)
using namespace std;
il int ci(){
	register char ch;int f=1;
	while(!isdigit(ch=getchar()))f=ch=='-'?-1:1;
	register int x=ch^'0';
	while(isdigit(ch=getchar()))x=(x*10)+(ch^'0');
	return f*x;
}

il ll gcd(ll a,ll b){ return b==0 ? a:gcd(b,a%b) ; }
ll a[100023];

int main(){
	int n=ci();
	ll M=ci();
	int o1=0, e1=0;
	rep(i,1,n) a[i]=ci()/2;
	ll g=a[1];
	rep(i,2,n){
		g=gcd(g,a[i]);
	}
	rep(i,2,n) a[i]/=g;
	int lcm=a[1];
	if( lcm&1 ) o1=1;
	else e1=1;
	rep(i,2,n){
		ll x=a[i];
		if( x&1 ) o1=1;
		else e1=1;
		lcm=lcm*x/gcd(lcm,x);
		if( lcm>M/g || (o1&&e1) ){
			printf("0\n");
			return 0;
		}
	}
	printf("%lld\n",(M-lcm)/(lcm*2)+1);
	return 0;
}