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: Runtime Error)

#include <assert.h>
#include <chrono>
#include <cstdio>
#include <iostream>
#include <limits.h>
#include <mutex>
#include <thread>
#include <vector>

long long evalmod(long long a, long long b){
  long long maxval = std::max(a, b);
  long long minval = std::min(a, b);
  if (minval==0){
    return maxval;
  }
  long long tmpval = a%b;
  return evalmod(b, tmpval);
}

int main(){
  int N;
  long long M;
  scanf("%d",&N);
  scanf("%lld",&M);
  std::vector<long long > a(N);
  for (int i=0; i<N; i++){
    scanf("%lld",&a[i]);
  }
  long long mod;
  long long temp = a[0];
  long long val = a[0];
  int countof2=0;
  while(temp%2==0){
    temp = temp/2;
    countof2++;
  }
  bool invalid = false;
  for (int i=1; i<N; i++){
    long long temp = a[i];
    int countof2_temp = 0;
    while(temp%2==0){
      countof2_temp++;
      temp = temp/2;
    }
    if (countof2!=countof2_temp){
      invalid = true;
    }
    mod = evalmod(a[i],val);
    val = (val*a[i])/mod;
    if (val>2*M){
      invalid = true;
    }
  }
  if (invalid){
    printf("0\n");
    return 0;
  }
  long long ans = (M+val*0.5)/val;
  //std::cout<<eval
  //printf("%lld\n",mod);
  //printf("%lld\n",val);
  printf("%lld\n",ans);
  return 0;
}