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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.ArrayList;
import java.util.HashSet;
import java.util.List;
import java.util.Scanner;
import java.util.Set;

public class Main {
  public static void main(String[] args) {
    Scanner scanner = new Scanner(System.in);

    int N = scanner.nextInt();
    long M = scanner.nextInt();
    Set<Integer> list = new HashSet<>();
    for (int i = 0; i < N; i++) list.add(scanner.nextInt() / 2);
    List<Integer> a = new ArrayList<>(list);

    int gcd = a.get(0);
    for (int num : a) gcd = gcd(gcd, num);

    long lcm = gcd;
    for (int num : a) lcm = lcm * (num / gcd);
    
    for (int num : a) {
      if ((lcm / num) % 2 == 0) {
        System.out.println(0);
        return;
      }
    }

    if (M < lcm) {
      System.out.println(0);
      return;
    }
    System.out.println((M - lcm) / (lcm * 2) + 1);
  }

  private static int countTwoPower(int a) {
    int cnt = 0;
    while (a % 2 == 0) {
      a /= 2;
      cnt++;
    }
    return cnt;
  }

  private static int gcd(int i, int j) {
    if (i > j) return gcd(j, i);
    while (i > 0) {
      int r = j % i;
      j = i;
      i = r;
    }
    return j;
  }
}