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 Kotlin (1.3.71) (judge verdict: Wrong Answer)

import java.io.PrintWriter

fun main(args: Array<String>) {
    PrintWriter(System.out).use { out ->
        solve(out)
        out.flush()
    }
}

fun solve(out: PrintWriter) {
    val (n, m) = readLongList()
    val aList = readLongList()

    // x = ak * (p + 0.5)
    // x = (ak / 2) * (2p + 1)
    // x = (ak / 2) * 奇数
    // a <= x <= m
    // xの個数
    var divCount = 0L
    for (a in aList) {
        var tmp = a
        var count = 0L
        while (tmp % 2 == 0L) {
            tmp /= 2
            count++
        }

        when {
            divCount == 0L -> divCount = count
            divCount != count -> {
                println(0)
                return
            }
        }
    }

    var t = aList.first() / 2
    for (a in aList) {
        t = lcm(t, a / 2)
    }

    var ans = 0L
    while (t * (2 * ans + 1) <= m) {
        ans++
    }
    println(ans)
}

/** O(log(n)): n = min(a, b) */
tailrec fun gcd(a: Long, b: Long): Long = if (b == 0L) a else gcd(b, a % b)

/** O(log(n)): n = min(a, b) */
fun lcm(a: Long, b: Long) = a / gcd(a, b) * b

fun readInteger() = readLine()!!.toInt()
fun readLong() = readLine()!!.toLong()
fun readStringList() = readLine()!!.split(" ")
fun readIntegerList() = readStringList().map(String::toInt)
fun readLongList() = readStringList().map(String::toLong)
fun readDoubleList() = readStringList().map(String::toDouble)