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++14 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include <cstdio>

using namespace std;

#define N 100000
long long int a[N];

int gcd(int a, int b)
{
    if (a < b) {
        a ^= b;
        b ^= a;
        a ^= b;
    }
    
    return b ? gcd(b, a % b) : a;
}

long long int calc_lcm(long long int a, long long int b)
{
    return a * b / gcd(a, b);
}

int main(void)
{
    int n;
    long long int m;
    scanf("%d", &n);
    scanf("%lld", &m);
    for(int i = 0; i < n; i++) {
        scanf("%lld", &a[i]);
    }

    long long int lcm = 1;
    for(int i = 0; i < n; i++) {
        lcm = calc_lcm(lcm, a[i]);
        if(lcm > m * 2) {
            printf("0\n");
            return 0;
        }
    }

    long long int ans = (m - lcm / 2) / lcm + 1;
    printf("%lld\n", ans);
    return 0;
}