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.*;

class Main {
    int N;
    long M;
    long[] A;

    public Main(){
        Scanner SC = new Scanner(System.in);
        N = SC.nextInt();
        M = SC.nextLong();
        A = new long[N];

        for(int i=0;i<N;i++){
            A[i] = SC.nextLong() / 2;
        }

        long valueLCM = valueLCM = LCM(A[0], A[1]);
        for(int i=1;i<N-1;i++){
            valueLCM = LCM(valueLCM, A[i+1]);
        }

        long L=-1, R=(M+1)/2;
        while(R-L > 1){
            long mid = L+(R-L)/2;

            if(valueLCM * (2*mid+1) > M)
                R = mid;
            else
                L = mid;
        }

        System.out.println(R);
    }

    long LCM(long val1, long val2){
        long maxValue = Math.max(val1, val2);
        long minValue = Math.min(val1, val2);

        if(minValue == 0) return maxValue;

        long val3 = maxValue * minValue;

        long temp = maxValue%minValue;
        while( temp != 0 ){
            maxValue = minValue;
            minValue = temp;
            temp=maxValue%minValue;
        }

        return val3/minValue;
    }

    public static void main(String[] args){
        Main m = new Main();
    }
}