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>

#define NREP() for(ll i = 0; i < n; i++)
#define REP(i, n) for(ll i = 0; i < n; i++)
#define ALL(x) (x).begin(), (x).end()
#define MSG(x) cout << x << endl;
#define IPT(x) ll t; cin >> t; x.push_back(t);
#define YN(x) x ? cout << "YES" << endl : cout << "NO" << endl;
#define Yn(x) x ? cout << "Yes" << endl : cout << "No" << endl;
#define yn(x) x ? cout << "yes" << endl : cout << "no" << endl;

using namespace std;
typedef long long ll;
typedef vector<ll> vl;
typedef vector<vector<ll>> vvl;
typedef vector<string> vs;
typedef vector<string> vvs;

const static ll MOD = 1e9 + 7;

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

ll lcm(ll a, ll b) {
    return a * b / gcd(a, b);
}

int main(int argc, char *argv[]) {
    ll n, m;
    cin >> n >> m;

    vl a;
    NREP() {
        IPT(a)
    }

    ll ans = 0;
    ll half_lcm = lcm(a.at(0), a.at(1));
    REP(i, n ) {
        if (i == 0 || i == 1) continue;
        half_lcm = lcm(a.at(i), half_lcm);
    }
    half_lcm /= 2;

    ll delta = half_lcm;
    while (half_lcm <= m) {
        ans++;
        half_lcm += delta * 2;
    }
    MSG(ans)
}