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<bits/stdc++.h>
using namespace std;
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
template<class T> using oset=tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
#define pb push_back
#define N 200001
#define ll long long
#define int ll 
#define pi pair<int , int>
#define pip pair<pair ,int>
#define mp make_pair 
#define f first
#define s second
#define mod 998244353
ll Mod(ll x, ll y, int p) 
{ll res = 1; x = x % p;
while (y > 0) 
{if(y & 1) res = (res*x) % p; 
y = y>>1;x = (x*x) % p; 
}return res; 
}
int* getlps(string pattern){
 int len = pattern.length();
 int * lps = new int[len];
 lps[0] = 0;
 int i = 1 , j = 0;
 while(i < len){
  if(pattern[i] == pattern[j]){
	  lps[i] = j+1;
   i++;
   j++;
  }
  else{
		if(j != 0){
		 j = lps[j-1];
		}
		else{
		 lps[i] = 0;
		 i++;
		}
  }
 }
 return lps;
}
class Triplet{
 public:
 int x ;
 int y ;
 int gcd;
};
Triplet extendedEuclid(int a , int b){
 if(b == 0){Triplet ans; ans.gcd = a;ans.x = 1;ans.y = 0;return ans;}
 Triplet smallAns = extendedEuclid(b , a%b);
 Triplet ans;
 ans.gcd = smallAns.gcd;
 ans.x = smallAns.y;
 ans.y = smallAns.x - (a/b)*smallAns.y;
 return ans;
}
int mmInverse(int a , int m){
 Triplet ans = extendedEuclid(a , m);
 return (ans.x+m)%m;
}
int fact[N];
void calfac(int n ){
fact[0] = 1;
for(int i = 1 ; i <= n+2; i++){
 fact[i] = (((fact[i-1]%mod)*(i%mod))%mod + mod)%mod;
}
 }
int calc(int n , int r){
 if(r > n)return 0;
 if(r == n)return 1;
 int ans = 1;
 ans = ((ans%mod)*(fact[n])%mod + mod)%mod;
 ans = ((ans%mod)*(mmInverse(fact[n-r] , mod)%mod) + mod)%mod;
 ans = ((ans%mod)*(mmInverse(fact[r] , mod)%mod) + mod)%mod;
 return (ans+mod)%mod;
}
int32_t main(){
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int n, upto;
cin>>n>>upto;
int arr[n];
for(int i = 0 ; i  < n;i++){
	cin>>arr[i];
	arr[i] = arr[i]/2;
}
int dp[n];
for(int i = 0 ; i <  n;i++){
 int val = arr[i];
 int cnt = 0;
 while(val%2 == 0){
  val=val/2;
  cnt++;
  if(val == 0)break;
 }
 dp[i] = cnt;
}
map<int , int> m;
int odd = 0;
for(int i = 0 ; i < n ;i++){
  if(dp[i] > 0){
   m[dp[i]]++;
  }
  else{
      odd++;
  }
}

if(((int)m.size()>=1 && odd != 0) || ((int)m.size() >= 2)){
   cout<<0<<endl;
   return 0;
}
// only odd values ,
int cnt = 0;
for(auto it:m)cnt = it.s;
//cout<<cnt<<endl;
int lcm = arr[0];
	for(int i = 1 ; i < n; i++){
		//arr[i] = 2*arr[i];
	     lcm = (lcm*arr[i])/(__gcd(lcm , arr[i]));
	}
	//for(int i = 0 ; i < cnt ; i++)lcm*=2;
	if(cnt)
	cout<<((upto/(lcm))+1)/2<<endl;
	else
    cout<<((upto/lcm)+1)/2<<endl;
   return 0;
	
}