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 9.2.1) (judge verdict: Accepted)

#include <stdio.h>

long long gcd(long long a, long long b)
{
	if (a > b) {
		a ^= b;
		b ^= a;
		a ^= b;
	}
	if (b % a == 0) return a;
	else return gcd(b % a, a);
}

int main()
{
	int i, N;
	long long M, a[100001];
	scanf("%d %lld", &N, &M);
	for (i = 1; i <= N; i++) scanf("%lld", &(a[i]));
	
	long long g = a[1], l, b;
	for (i = 2; i <= N; i++) g = gcd(g, a[i]);
	for (b = 2; (g & b) == 0; b <<= 1);
	for (i = 1, l = b >> 1; i <= N; i++) {
		a[i] /= b;
		if (a[i] % 2 == 0) break;
		l *= a[i] / gcd(a[i], l);
		if (l > M) break;
	}
	
	if (i <= N) printf("0\n");
	else printf("%lld\n", (M + l) / l / 2);
	fflush(stdout);
	return 0;
}