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 (GCC 5.4.1) (judge verdict: Accepted)

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<stdlib.h>
#include<limits.h>

#define rep(i,begin,end) for(int i=begin; i<end; i++)
#define lld long long int

long long int gcd(long long int a, long long int b){
    long long int c;
    if(a<b){
        c = a;
        a = b;
        b = c;
    }
    while(a%b!=0){
        c=a%b;
        a=b;
        b=c;
    }
    return b;
}

int main(){
    int n;
    lld m;
    scanf("%d %lld", &n, &m);
    lld a[n];
    rep(i, 0, n){
        scanf("%lld", &a[i]);
        a[i] /= 2;
    }
    lld x=1, y;
    int cnt = 0, f = 0;
    int buf = a[0];
    while(buf%2 == 0){
        buf /= 2;
        cnt++;
    }
    rep(i, 1, n){
        buf = a[i];
        int c = 0;
        while(buf%2 == 0){
            buf /= 2;
            c++;
        }
        if(cnt != c){
            f = 1;
        }
    }
    rep(i, 0, n){
        y = x*a[i];
        x = y / gcd(x, a[i]);
    }
    if(f){
        printf("0");
    }else{
        printf("%lld", (m/x + 1)/2);
    }

    return 0;
}