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: Accepted)

#include<bits/stdc++.h>
using namespace std;
using ll = long long;
#define REP(i, a, b) for(int i = a, i##_len = (b); i < i##_len; ++i)

const int MAX_N = 100100;
ll A[MAX_N], B[MAX_N];
ll N, M;

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()
{
    scanf("%lld %lld", &N, &M);
    for(int i = 0; i < N; ++i) scanf("%lld", &A[i]);
    for(int i = 0; i < N; ++i) B[i] = A[i]/2LL;
    ll LCM1 = 1LL;
    for(int i = 0; i < N; ++i) LCM1 = lcm(LCM1, B[i]);
    ll LCM2 = 2LL*LCM1;
    //printf("%lld %lld\n", LCM1, LCM2);
    for(int i = 0; i < N; ++i) if((LCM1/B[i])%2==0) { printf("0\n"); return 0;}
    if(LCM1 > M ) { printf("0\n"); return 0;}
    ll res = 1LL + (M - LCM1)/LCM2;
    printf("%lld\n", res);
    return 0;
}