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.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Scanner;

/**
 * Built using CHelper plug-in Actual solution is at the top
 *
 * @author Silviase
 */
public class Main {

  public static void main(String[] args) {
    InputStream inputStream = System.in;
    OutputStream outputStream = System.out;
    Scanner in = new Scanner(inputStream);
    PrintWriter out = new PrintWriter(outputStream);
    ABC150D solver = new ABC150D();
    solver.solve(1, in, out);
    out.close();
  }

  static class ABC150D {

    public void solve(int testNumber, Scanner in, PrintWriter out) {
      int n = in.nextInt();
      int m = in.nextInt();
      int[] ar = new int[n];
      boolean f = true;
      int cnt2 = 0;
      for (int i = 0; i < n; i++) {
        ar[i] = in.nextInt();
      }

      while (true) {
        for (int i = 0; i < n; i++) {
          if (ar[i] % 2 != ar[0] % 2) {
            // 素因数2の個数が異なる場合はNG
            f = false;
          }
        }
        if (!f) {
          break;
        }
        if (ar[0] % 2 == 0) {
          cnt2++;
          for (int i = 0; i < n; i++) {
            ar[i] /= 2;
          }
        } else {
          break;
        }
      }

      if (!f) {
        out.println(0);
      } else {
        long lcm = arrayLcm(ar, m);
        if (lcm == -1) {
          out.println(0);
        } else {
          out.println((m - lcm) / ((long) (Math.pow(2, cnt2) * lcm)) + 1);
        }
      }

    }

    long arrayLcm(int[] ar, int m) {
      long lcm = ar[0];
      for (int i = 1; i < ar.length; i++) {
        lcm = lcm * (long) ar[i] / gcd(lcm, ar[i]);
        if (lcm > m) {
          return -1;
        }
      }
      return lcm;
    }

    long gcd(long a, long b) {
      return b == 0 ? a : gcd(b, a % b);
    }

  }
}