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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.io.*;
import java.util.*;
public class Main {

	public static void main(String[] args) {
		FastScanner sc = new FastScanner();
//		StringBuilder out = new StringBuilder();
		int n = sc.nextInt();
		long m = sc.nextInt();
		int[] arr = new int[n];
//		long g = 0;
		for(int i = 0; i < n; i++) {
			arr[i] = sc.nextInt()/2;
//			g = gcd(g, arr[i]);
		}
		long mcd = 1;
		for(int k: arr) {
			mcd = mcd(mcd, k);
			if(mcd > m) break;
		}
		if(mcd <= m) System.out.println((m-mcd)/(2*mcd) + 1);
		else System.out.println(0);
	}
	static long mcd(long a, long b) {
		return a * b / gcd(a, b);
	}
	static long gcd(long a, long b) {
		if(b > a) return gcd(b, a);
		else if(b == 0) return a;
		else return gcd(b, a % b);
	}
	static int calc(int[] arr, int n) {
		int res = 0;
		boolean[] seen = new boolean[n];
		for(int i = 0; i < n-1; i++) {
			int k = 0;
			for(int j = 0; j < n; j++) {
				if(!seen[j] && j < arr[i]) k++;
			}
			seen[arr[i]] = true;
			res += k;
			res *= (n-i-1);
		}
		return res;
	}

	static class FastScanner {
		public BufferedReader reader;
		public StringTokenizer tokenizer;
		public FastScanner() {
			reader = new BufferedReader(new InputStreamReader(System.in), 32768);
			tokenizer = null;
		}
		public String next() {
			while (tokenizer == null || !tokenizer.hasMoreTokens()) {
				try {
					tokenizer = new StringTokenizer(reader.readLine());
				} catch (IOException e) {
					throw new RuntimeException(e);
				}
			}
			return tokenizer.nextToken();
		}
		public int nextInt() {
			return Integer.parseInt(next());
		}
		public long nextLong() {
			return Long.parseLong(next());
		}
		public double nextDouble() {
			return Double.parseDouble(next());
		}
		public String nextLine() {
			try {
				return reader.readLine();
			} catch(IOException e) {
				throw new RuntimeException(e);
			}
		}
	}

}