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 <iostream>
#include <vector>
using namespace std;

long long GCD(long long x, long long y) {
    if ( y == 0) return x;
    else return GCD(y, x % y);
}

long long solve(int N, long long M, vector<long long> &a) {
    // 2 で割れるだけ割る
    bool same = true;
    while (a[0] % 2 == 0) {
        for (int i = 0; i < N; ++i) {
            if (a[i] % 2 != 0) return 0;
            a[i] /= 2;
        }
        M /= 2;
        for (int i = 0; i < N; ++i) if (a[i] % 2 == 0) return 0;
    }

    // lcm
    long long lcm = 1;
    for (int i = 0; i < N; ++i) {
        long long g = GCD(lcm, a[i]);
        lcm = lcm / g * a[i];
        if (lcm > M) return 0;
    }
    return (M / lcm + 1) / 2;
}

int main() {
    int N; long long M;
    cin >> N >> M;
    vector<long long> a(N);

    // あらかじめ a は一回 2 で割っておく
    for (int i = 0; i < N; ++i) cin >> a[i], a[i] /= 2;
    cout << solve(N, M, a) << endl;
}