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>
#include <stdlib.h>
#include <string.h>

unsigned long long gcd(unsigned long long x, unsigned long long y){
    return y == 0 ? x : gcd(y, x % y);
}

unsigned long long lcm(unsigned long long x,unsigned long long y){
    return x / gcd(x,y) * y;
}

int main(){
    unsigned long long i,n,m,a[100002]={},total=0,p=0;
    scanf("%lld %lld",&n,&m);
    scanf("%lld",&a[0]);
    for (i=1; i<n; i++) scanf(" %lld",&a[i]);
    for (i=0; i<n; i++) a[i]/=2;
    total=lcm(a[0], a[1]);
    for (i=2; i<n; i++) total=lcm(total, a[i]);
    for(i = 0; i < n; i++)
        if(total / a[i] % 2 == 0){
            printf("0\n");
            return 0;
        }
    printf("%lld\n",(m/total+1)/2);
    return 0;
}