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 C (Clang 3.8.0) (judge verdict: Runtime Error)

#include <stdio.h>
#include <math.h>
int a[50001];
int main()
{
    int n,m, i,num;
    int mingcd = 1;
    int count = 0;
    scanf("%d%d",&n, &m);
    for(i = 0; i < n; i++){
        scanf("%d",&num);
        if(!a[num/2]){
            a[num/2]++;
            mingcd *= num/2;
        }
    }
    for(i = 1; mingcd * i <= m; i++)
        if(mingcd * i >= 1 && mingcd * i <= m)   count++;
     printf("%d\n",count/2+1);

    return 0;
}