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: Wrong Answer)

#include <algorithm>
#include <cmath>
#include <iomanip>
#include <iostream>
#include <limits>
#include <map>
#include <queue>
#include <set>
#include <tuple>
#include <vector>

using namespace std;

#define rep(i, n) for (int64_t i = 0; i < (n); i++)
#define irep(i, n) for (int64_t i = 0; i <= (n); i++)
#define rrep(i, n) for (int64_t i = (n)-1; i >= 0; i--)
#define rirep(i, n) for (int64_t i = n; i >= 0; i--)

int64_t dceil(int64_t, int64_t);
int64_t dfloor(int64_t, int64_t);

int64_t dceil(int64_t x, int64_t y)
{
    if (y < 0) {
        return dceil(-x, -y);
    }
    if (x < 0) {
        return -dfloor(-x, y);
    }
    return (x + y - 1) / y;
}

int64_t dfloor(int64_t x, int64_t y)
{
    if (y < 0) {
        return dfloor(-x, -y);
    }
    if (x < 0) {
        return -dceil(-x, y);
    }
    return x / y;
}

int64_t gcd(int64_t x, int64_t y)
{
    if (x == 0) {
        return y;
    }
    return gcd(y % x, x);
}

int main()
{
    int64_t n, m;
    cin >> n >> m;

    int64_t L = 1;
    vector<int64_t> a(n);
    rep(i, n)
    {
        cin >> a[i];
        a[i] /= 2;
        L = L / gcd(L, a[i]) * a[i];
    }

    if (L % 2 == 0) {
        cout << 0 << endl;
    } else {
        int64_t lb = dceil(1 - L, 2 * L), ub = dfloor(m - L, 2 * L);
        cout << max(0L, ub - lb + 1) << endl;
    }

    return 0;
}