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 <iostream>
#include <vector>
#include <algorithm>
#include <map>
#include <string>
#include <cmath>

using namespace std;

#define ll long long

int main(void){
    int n, m;
    cin >> n >> m;
    vector<int> a(n);
    int max_num=0;
    for(int i=0;i < n;++i){
        cin >> a[i];
        max_num = max(a[i], max_num);
    }

    vector<int> cand(m, 0);
    int tail;
    for(int p = 1;p*max_num+max_num/2 <= m; ++p){
        cand[p-1] = max_num*p + max_num/2;
        tail = p;
    }

    for(int i = 0;i < n;++i){
        for(int j = 0;j < tail;++j){
            if( (cand[j]-a[i]/2) % a[i] != 0){
                cand[j] = cand[tail-1];
                tail--;
                j--;
            }
        }
    }

    cout << tail << endl;




    return 0;
}