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>
//#include <stdlib.h>
//#include <string.h>
//#include <stdbool.h>
//#include <limits.h>
//#include <math.h>

int gcd(int a, int b) {
  if (b == 0) return a;
  return gcd(b, a%b);
}

int lcm(int a, int b) {
  int g = gcd(a, b);
  return a*b/g;
}

int main() {
  int n, m;
  scanf("%d %d", &n, &m);
  int a[n];
  scanf("%d", a);
  int b = a[0], cnt = 0, l = b/2;
  while (b%2 == 0) {
    cnt++;
    b /= 2;
  }
  for (int i=1; i<n; i++) {
    scanf("%d", a+i);
    int cnti = 0;
    b = a[i];
    while (b%2 == 0) {
      cnti++;
      b /= 2;
    }
    if (cnt != cnti) {
      printf("0\n");
      return 0;
    }
    l = lcm(a[i]/2, l);
  }
  printf("%d\n", m/l-m/(2*l));
  return 0;
}