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>

using namespace std;

#define pb push_back
#define mp make_pair
#define ff first
#define ss second
#define BIT(X,i) ((X>>i)&1)
#define REP(i,a,b) for(int i = (a); i < (b); ++i)
#define rep(i,n) for(int i = 0; i < (n); ++i)
#define all(a) (a).begin(), (a).end()
#define rall(a) (a).rbegin(), (a).rend()
#define SORT(a) sort(all(a))
#define RSORT(a) sort(rall(a))
#define ina(x,n) rep(x_i,n) cin>>x[x_i]
#define outa(x,n) rep(x_i,n) cerr<<x[x_i]<<" \n"[x_i==n-1]

typedef long long ll;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef pair<ll, ll> pll;
typedef vector<pll> vll;

const int MOD = 1e9+7;
const int INF = (1<<28);
const ll INFL = (1LL<<60);

ll gcd(ll a, ll b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}

ll lcm(ll a, ll b) {
  ll g = gcd(a, b);
  return a / g * b;
}

int main() {
  ll n, m, g = 1; scanf("%lld %lld", &n, &m);
  bool flag = false;
  rep(i,n) {
    ll a;
    scanf("%lld", &a);
    g = lcm(g, a/2);
    if (g > m) flag = true;
  }
  if (flag) {
    cout << 0 << endl;
    return 0;
  }
  ll ans = max(0.0, floor((double)(m+g)/2/g));
  printf("%lld\n", ans);
  return 0;
}