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: Runtime Error)

#include<stdio.h>
#include<stdlib.h>

int compare_int_big(const void *a,const void *b){
    return *(int *)b - *(int *)a;
}

int compare_int_small(const void *a,const void *b){
    return *(int *)a - *(int *)b;
}

int main(void)
{
    int ans = 0;
    int kouho = 0;
    int saishouKBS = 0;
    int flag = 0;

    int N,M;
    int A[100010];

    scanf("%d",&N);
    scanf("%d",&M);

    for (int i = 0; i < N; i++)
    {
        scanf("%d",&A[i]);
    }
    
    qsort(A,N,sizeof(int),compare_int_big);

    for(int i = 0; kouho <= M ;i++){
        kouho = A[0]*i+A[0]/2;
        if(kouho > M) break;
        flag = 1;
        for(int j = 1; j < N ; j++){
            if(kouho%A[j] == A[j]/2){

            }else{
                flag = 0;
            }
        }
        if(flag == 1){
            ans++;
            saishouKBS = kouho;
            break;
        }
    }

    ans = M/saishouKBS;
    ans /= 2;
    ans ++;

    printf("%d",ans);

    return 0;
}