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>
#include<stdbool.h>
#include<string.h>
#include<math.h>
#include<inttypes.h>
#define mprint(x) printf(FORMATCONVERSION(x),(x))
#define FORMATCONVERSION(x) _Generic((x),\
                                    int:"%d",unsigned int:"%u",\
                                    double:"%lf",long long int:"%lld",\
                                    unsigned long long int:"%llu",\
                                    unsigned long : "%lu",\
                                    long int : "%ld",\
                                    char :"%c",char*:"%s")


uint32_t gcd(uint32_t a, uint32_t b) {
  uint32_t v0 = a, v1 = b, v2 = a % b;
  while (v2 != 0) {
    v0 = v1;
    v1 = v2;
    v2 = v0 % v1;
  };
  return v1;
}

uint32_t lcm(uint32_t a, uint32_t b) {
  return a * b / gcd(a, b);
}

int main(void){
    int n;
    uint32_t m;
    scanf("%d %"SCNu32,&n,&m);
    uint32_t array[n];
    for(int i = 0;i < n;i++){
        scanf("%" SCNu32,&array[i]);
    }
    if(n == 1){
        int ans = (((m * 2) / array[0]) + 1) / 2;
        mprint(ans);
        return 0;
    }
    uint64_t g_num = lcm(array[0],array[1]);
    for(int i = 2;i < n;i++){
        g_num = lcm(g_num,array[i]);
    }

    int ans = (((m * 2) / g_num) + 1) / 2;
    mprint(ans);
    return 0;
}