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: Accepted)

#include<stdio.h>
#define ll long long

ll gcd(ll a,ll b){
  if(b==0) return a;
  return gcd(b,a%b);
}

int main(){
  ll n,m,i,a[100005];
  scanf("%lld %lld",&n,&m);
  for(i=0;i<n;i++){
    scanf("%lld", &a[i]);
    a[i] /= 2;
  }
  ll l=a[0];
  for(i=0;i<n;i++) l *= a[i]/gcd(l,a[i]); // l = (lとa[i]の最小公倍数) => (最終的な l) = (全ての最小公倍数)
  for(i=0; i<n; i++){
    if(l/a[i]%2==0){
      printf("0\n");
      return 0;
    }
  }
  printf("%lld\n",(m/l+1)/2);//奇数である公倍数の数 *テクニック: (x+1)/2は繰り上がり
  return 0;
}