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 (Clang 3.8.0) (judge verdict: Accepted)


#include <cassert>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <vector>
using namespace std;
#define REP(i, n) for(int i = 0; i < (int)(n); ++i)
typedef long long ll;

int n, m;
ll vs[100000+10];
ll ts[100000+10];


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

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


ll solve(void) {
  REP(i, n) {
    while(vs[i] % 2 == 0) {
      ts[i]++;
      vs[i] /= 2;
    }
  }
  REP(i, n) {
    if(ts[0] != ts[i]) {
      return 0;
    }
  }
  ll llcm = 1;
  REP(i, ts[0]) {
    llcm *= 2;
  }
  REP(i, n) {
    llcm = lcm(llcm, vs[i]);
    if(llcm > m) {
      return 0;
    }
  }
  return (m / llcm + 1) / 2;
}

int main(void) {
  scanf("%d%d", &n, &m);
  REP(i, n) {
    scanf("%lld", &vs[i]);
    vs[i] /= 2;
  }

  ll res = solve();
  cout << res << endl;
  
  return 0;
}