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)

import java.util.*;

public class Main {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int m = sc.nextInt();
        int[] a = new int[n];
        boolean existX = true;
        int prevFact2Cnt = 0;
        for(int i = 0; i < n; ++i) {
            a[i] = sc.nextInt() >> 1;
            if(i == 0) {
                prevFact2Cnt = countFactor(a[i], 2);
                existX = true;
            } else if(existX) {
                int tmpFact2Cnt = countFactor(a[i], 2);
                existX = (prevFact2Cnt == tmpFact2Cnt);
                prevFact2Cnt = tmpFact2Cnt;
            }
        }
        
        if(!existX) {
            System.out.println(0);
            return;
        }
        
        long minX = findArrayLcm(a, n);
        long q = m / minX;
        long ans = q >> 1;
        if(q % 2 == 1) ++ans;
        System.out.println(ans);
    }
    
    static int countFactor(int num, int factor) {
        int res = 0;
        while(num % 2 == 0) {
            ++res;
            num /= factor;
        }
        return res;
    }
    
    static long findGcd(long a, long b) {
        if(a < b) return findGcd(b, a);
        if(b == 0) return a;
        return findGcd(b, a % b);
    }
    
    static long findLcm(long a, long b) {
        long gcd = findGcd(a, b);
        long res = a / gcd * b;
        return res;
    }
    
    static long findArrayLcm(int[] ar, int size) {
        long res = ar[0];
        for(int i = 1; i < size; ++i) res = findLcm(res, ar[i]);
        return res;
    }
}