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 (Clang 3.8.0) (judge verdict: Accepted)

#include <stdio.h>
#include <stdint.h>
#include <inttypes.h>

int nd(uint64_t x)
{
    int d;
    for(d=0;x%2==0;x/=2,d++);
    return d;
}

uint64_t gcd(uint64_t x, uint64_t y)
{
    uint64_t z;
    while(y != 0){
        z = x % y;
        x = y;
        y = z;
    }
    return x;
}

int main()
{
    int N, M, a;
    scanf("%d%d", &N, &M);
    scanf("%d", &a);

    int d = nd(a);
    if(d==0){
        printf("%d\n", 0);
        return 0;
    }
    uint64_t h = a/2;
    for(int ni=1; ni<N; ni++){
        int an;
        scanf("%d", &an);
        if(d != nd(an)){
            printf("%d\n", 0);
            return 0;
        }
        h *= (an/2) / gcd(h, (an/2));
    }

    printf("%"PRIu64,  M/h - M/(2*h));
    
    return 0;
}