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 (GCC 5.4.1) (judge verdict: Runtime Error)

#include <stdio.h>

long euclid(long a, long b)
{
    long temp;

    if( a < b ) { temp = a; a = b; b = temp; }
    if( b < 1 ) return -1;

    if( a % b == 0 ) return b;
    return euclid( b, a % b );
}

long lcm(long n, long m) {

  return (n * m) / euclid(n, m);

}

long lcma(long a[], int len) {
  int i;
  long ansl;
  ansl = a[0];
  for(i = 1; i < len; i++) {
    ansl = lcm(ansl, a[i]);
  }
  return ansl;
}
	 
int main(void) {
  int i, n;
  long m;
  long x;
  double y;
  long z;
  long a[100001];

  scanf("%d %ld", &n, &m);
  for(i = 0; i < n; i++) {
    scanf("%ld", a + i);
  }

  x = lcma(a, n);
  y = (double) m / (double) x;
  z = y + 0.5;
  
  printf("%ld\n", z);

  return 0;
}