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: Time Limit Exceeded)

#include <bits/stdc++.h>
typedef long long ll;
using namespace std;
ll MOD = 1e9+7;

#define vec     vector<int>
#define vecll   vector<ll>
#define vec2    vector<vector<int>>
#define vec2ll  vector<vector<ll>>
#define vpair   vector<pair<int, int>>
#define vpairll vector<pair<ll, ll>>

#define rep(i,n)         for(ll i=(ll)0; i<(ll)n; i++)
#define REP(i,m,n)       for(ll i=(ll)m; i<(ll)n; i++)
#define arr(var, n)      vec var(n); rep(i,n){cin >> var[i];}
#define arrll(var, n)    vecll var(n); rep(i,n){cin >> var[i];}
#define all(var)         (var).begin(), (var).end()
#define sortall(var)     sort(all(var))
#define mpair(v1, v2)    make_pair(v1, v2)
#define pb(var)          push_back(var)
#define prt(var)         cout << var << "\n"
#define scn(var)         cin >> var;
#define pYes()           prt("Yes")
#define pNo()            prt("No")
#define prtd(n, var)     cout << fixed << setprecision(n) << var << "\n"
#define prt2(v1, v2)     cout << v1 << " " << v2 << "\n"
#define prt3(v1, v2, v3) cout << v1 << " " << v2 << " " << v3 << "\n"
#define prtall(var)      rep(i,var.size()){cout << var[i] << " ";} cout << "\n"
//----------------------------------------------------------------
ll gcd(ll x, ll y){
  if(x<y){return gcd(y,x);}
  if(y==0 || x==y) {return x;}
  else {return gcd(y, x%y);}
}

ll lcm(ll x, ll y){
  return x*y/gcd(x, y);
}

ll div2(ll x){
  ll num=0;
  while(x%2==0){
    x /= 2;
    num++;
  }
  return num;
}

int main(void) {
  ll n, m;
  cin >> n >> m;
  arrll(a,n);
  rep(i,n){
    a[i] /= 2;
  }

  ll k=div2(a[0]);
  rep(i,n){
    ll p=div2(a[i]);
    if(k!=p){
      prt(0); return 0;
    }
  }

  ll tmp=1;
  rep(i,n){
    tmp = lcm(a[i], tmp);
  }

  ll ans=0;
  if(tmp>m){
    ans=0;
  } else {
    ans=1+(m-tmp)/(2*tmp);
  }


  prt(ans);
}