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: Runtime Error)

package com.company;

import java.io.*;
import java.util.*;

public class Main {
    public static Scanner sc = new Scanner(System.in);
    public static PrintWriter out = new PrintWriter(System.out, true);
    public static long n, m;
    public static long[] a;

    public static long gcd(long a, long b) {
        while (b > 0) {
            long c = a % b;
            a = b;
            b = c;
        }
        return a;
    }

    public static long lcm(long a, long b) {
        return a * b / gcd(a, b);
    }

    public static void Input() {
        n = sc.nextLong();
        m = sc.nextLong();
        a = new long[(int) n];
        for (int i = 0; i < n; i++) {
            a[i] = sc.nextLong();
            a[i] /= 2;
        }
    }

    public static void Solve() {
        long x = a[0];
        for (int i = 1; i < n; i++) {
            x = lcm(x, a[i]);
            if (x > m) {
                out.println(0);
                return;
            }
        }

        for (int i = 0; i < n; i++)
            if ((x / a[i]) % 2 == 0) {
                out.println(0);
                return;
            }

        long res = (m / x - 1) / 2 + 1;
        out.println(res);
    }

    public static void main(String[] args) {
        Input();
        Solve();
    }
}