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;
using lint = long long int;
#define FOR(i, begin, end) for(int i=(begin),i##_end_=(end);i<i##_end_;i++)
#define IFOR(i, begin, end) for(int i=(end)-1,i##_begin_=(begin);i>=i##_begin_;i--)
#define REP(i, n) FOR(i,0,n)
#define IREP(i, n) IFOR(i,0,n)
constexpr int MOD = 1000000007;
constexpr int INF = 2147483647;
lint gcd(lint a, lint b) {
  if(a < b) {
    swap(a, b);
  }
  while(b > 0) {
    lint r = a % b;
    a = b;
    b = r;
  }
  return a;
}
lint lcm(lint a, lint b) {
    return a * b / gcd(a, b);
}
int main()
{
    lint N, M;
    cin >> N >> M;
    vector<lint> A(N);
    REP(i, N) cin >> A[i];
    bool flg1 = false;
    bool flg2 = false;
    vector<lint> div(N, 1);
    REP(i, N) {
        while(A[i]%2 == 0) {
            A[i] /= 2;
            div[i] *= 2;
        }
    }
    FOR(i, 1, N) {
        if(div[i] != div[i-1]) {
            cout << 0 << endl;
            return 0;
        }
    }
    lint alllcm = A[0];
    FOR(i, 1, N){
        alllcm = lcm(alllcm, A[i]);
        if(alllcm > M) {
            cout << 0 << endl;
            return 0;
        }
    }
    cout << (M / alllcm + 1) / 2 << endl;
}