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: Wrong Answer)


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

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

int main(void){
	int n, m, i;
	long l=1;
	scanf("%d%d", &n, &m);
	for(i=0; i<n; i++){
		int a;
		scanf("%d", &a);
		a/=2;
		l = l/gcd(l, a)*a;
		if(l>m) break;
	}
	if(i<n){
		puts("0");
		return 0;
	}
	printf("%d\n", m/l - m/(l*2));
}