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 Go (1.6) (judge verdict: Accepted)

package main

import "fmt"

func FactoredPrimeNumber(num int) map[int]int {
	return factor(num, 2)
}

func merge(m1, m2 map[int]int) map[int]int {
	ans := map[int]int{}
	for k, v := range m1 {
		ans[k] += v
	}
	for k, v := range m2 {
		ans[k] += v
	}
	return ans
}

func factor(num, pnum int) map[int]int {
	result := make(map[int]int)
	if pnum*pnum > num {
		if num != 1 {
			result[num] += 1
		}
		return result
	}

	if num%pnum == 0 {
		num /= pnum
		result[pnum]++
	} else {
		pnum++
	}
	return merge(result, factor(num, pnum))
}

func count(n int) int {
	cnt := 0
	for n%2 == 0 {
		cnt++
		n /= 2
	}
	return cnt
}

func gcd(x, y int) int {
	if y == 0 {
		return x
	}
	return gcd(y, x%y)
}
func main() {
	var N, M int
	fmt.Scanf("%d %d\n", &N, &M)

	a := make([]int, N)
	for i := 0; i < N; i++ {
		var ai int
		fmt.Scan(&ai)
		a[i] = ai / 2
	}
	cnt := count(a[0])
	M = M >> uint(cnt)
	for i := 0; i < N; i++ {
		if cnt != count(a[i]) {
			fmt.Println(0)
			return
		}
		a[i] = a[i] >> uint(cnt)
	}
	lcm := 1
	for i := 0; i < N; i++ {
		r := gcd(lcm, a[i])
		lcm = lcm / r * a[i]
	}
	if lcm > M {
		fmt.Println(0)
		return
	}
	fmt.Println((M/lcm + 1) / 2)
}