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: Accepted)


import java.util.Scanner;

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

        final long[] array = new long[n];
        for (int i = 0; i < n; i++) {
            array[i] = scanner.nextLong() / 2;
        }

        long base = array[0];
        for (int i = 1; i < n; i++) {
            final long gcd = computeGCD(base, array[i]);
            if (base / gcd % 2 == 0 || array[i] / gcd % 2 == 0) {
                System.out.println(0);
                return;
            }
            base = base * array[i] / gcd;

            if (base > m) {
                System.out.println(0);
                return;
            }
        }

        System.out.println((m / base + 1) / 2);
    }

    private static long computeGCD(final long a, final long b) {
        if (b > a) {
            return computeGCD(b, a);
        }

        if (b == 0) {
            return a;
        }

        return computeGCD(b, a % b);
    }
}