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 <math.h>
int minimam(int *a,int n){
  int min=*a;
  for(int i=0;i<n;i++){
    if(*(a+i)<min){
      min=*(a+i);
    }else{}
  }
  return min;
}

int main(){
  int n,m,min,counta=0;
  scanf("%d %d",&n,&m);
  int a[n];
  for(int i=0;i<n;i++){
    scanf("%d",&a[i]);
  }
  min=minimam(a,n);
  
  for(int i=m;i>min;i--){
    int count=0;
    for(int j=0;j<n;j++){
      if((2*i/a[j])%2==1 && (2*i)%a[j]==0){}else{
        break;
      }
      count++;
    }
    if(count==n){
      counta++;
    }
  }
  printf("%d",counta);
  return 0;
}