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>
using namespace std;

#define REP(i, n) for (ll (i) = 0 ; (i) < (ll)(n) ; ++(i))
#define REPN(i, m, n) for (ll (i) = m ; (i) < (ll)(n) ; ++(i))
#define REP_REV(i, n) for (ll (i) = (ll)(n) - 1 ; (i) >= 0 ; --(i))
#define REPN_REV(i, m, n) for (ll (i) = (ll)(n) - 1 ; (i) >= m ; --(i))

#define INF 2e9
#define INF_LL 1LL<<60
#define ll long long

typedef pair<ll, ll> P;

ll gcd(ll a, ll b){
    ll tmp;
    if(a<b){
        tmp = a;
        a = b;
        b = tmp;
    }

    ll r = a%b;
    while(r!=0){
        a = b;
        b = r;
        r = a % b;
    }

    return b;
}

ll kaisuu(ll a) {
    int i = 0;
    while(1) {
        if(a % 2 == 0) {
            i++;
            a /= 2;
        } else {
            break;
        }
    }

    return i;
}

set<ll> st;

int main() {
    ll n, m, ans = 0;
    cin >> n >> m;
    ll kobai, koyaku;
    int kaisu = 0;

    REP(i, n) {
        ll tmp;
        cin >> tmp;
        tmp /= 2;
        st.insert(tmp);
    }

    kaisu = kaisuu(*st.begin());
    for(auto s: st) {
        cout << kaisu << endl;
        if(kaisu != kaisuu(s)) {
            cout << "0" << endl;
            return 0;
        }
    }

    koyaku = kobai = *st.begin();
    for(auto a: st) {
        koyaku = gcd(kobai, a);
    }

    kobai = 1;
    for(auto a: st) {
        if(kobai % (a/koyaku) != 0) kobai *= a / koyaku;
    }

    if(kobai > m) {
        cout << "0" << endl;
        return 0;
    }

    //cout << kobai << endl;

    ans = (m - kobai) / (2*kobai) + 1;

    cout << ans << endl;


    return 0;
}