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: Wrong Answer)

#include <stdio.h>
long long int N, M, a[100001], sum;
long long int GCD(long long int a, long long int b)
{
	while(b^=a^=b^=a%=b);
	return a;
}
long long int LCM(long long int a, long long int b)
{
	return a*b/GCD(a, b);
}
int main()
{
    scanf("%lld%lld", &N, &M);
    M *= 2;
    for(int i=1;i<=N;i++)
    {
    	scanf("%lld", &a[i]);
	}
	int lcm = 1;
	for(int i=1;i<=N;i++)
	{
		if(lcm!=1&&lcm%a[i]==0)
		    continue;
		lcm = LCM(lcm, a[i]);
	}
//	printf("%lld", a[1]);
	long long int temp = lcm;
	while(lcm<=M)
	{
		if((lcm/a[1])%2==1)
		    sum++;
		lcm += temp;
	}
	printf("%lld\n", sum);
	return 0;
}