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: Accepted)

#include <stdio.h>

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

int main(void) {
	int n, m, a, i;
	scanf("%d%d", &n, &m);
	scanf("%d", &a);
	long lcm = a/2;
	for(i = 0; i < n - 1; i++) {
		scanf("%d", &a);
		int gcd = getgcd((int)lcm, a/2);
		if((lcm/gcd)%2 == 0 || (a/gcd)%4 == 0) {
			printf("0\n");
			return 0;
		}
		lcm = lcm*a/(2*gcd);
		if(m < lcm) {
			printf("0\n");
			return 0;
		}
	}
	printf("%d\n", (int)((m + lcm)/(2*lcm)));
	return 0;
}