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: Runtime Error)

#include <bits/stdc++.h>
using namespace std;

int f(int x) {
  int ans = 0;
  while(x % 2 == 0) {
    ans++;
    x /= 2;
  }
  return ans;
}

int g(int x, int y) {
  if(y % x == 0) {
    return x;
  }
  else {
    return g(y % x, x);
  }
}

int main() {
  int N, M;
  cin >> N >> M;
  vector<int> b(N);
  for(int i = 0; i < N; i++) {
    int a;
    cin >> a;
    b.at(i) = a / 2;
  }
  vector<int> count(N);
  for(int i = 0; i < N; i++) {
    count.at(i) = f(b.at(i));
  }
  sort(count.begin(), count.end());
  if(count.at(0) != count.at(N - 1)) {
    cout << 0 << endl;
  }
  else {
    for(int i = 0; i < N; i++) {
      b.at(i) /= pow(2, count.at(i));
    }
    int a = b.at(0);
    for(int i = 1; i < N; i++) {
      int m = g(min(a, b.at(i)), max(a, b.at(i)));
      a = a * b.at(i) / m;
    }
    int x = a * pow(2, count.at(0));
    cout << (M / x - 1) / 2 + 1 << endl;
  }
}