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 (GCC 5.4.1) (judge verdict: Runtime Error)

#include<bits/stdc++.h>
#pragma GCC optimize("Ofast")
using namespace std;
long long gcd(long a,long b){
    if(a>b){swap(a,b);}
    if(a==0){return b;}
    else{return gcd(b%a,a);}
}

long long ord2(long long n){
    long long ret =0;
    while(n%2==0){
        n/=2;
        ret++;
    }
    return ret;
}
signed main(){
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    long long n,m,ans=0,L=1,ord,d;
    cin>>n>>m;
    vector<long long>a(n);
    for(int i=0;i<n;i++){
        cin>>a[i];
    }
    ord=ord2(a[0]);
    d=1<<ord;
    for(int i=0;i<n;i++){
        if(ord!=ord2(a[i])){cout<<0;return 0;}
        a[i]/=d;
        L=(L/gcd(L,a[i]))*a[i];
    }
    L*=d/2;
    if(L>m){cout<<0;return 0;}
    for(long long i=L;i<=m;i+=2*L){
        ans++;
    }
    cout<<ans;
    return 0;
}