Problem: 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).

Examples

Input

2 50
6 10


Output

2


Input

3 100
14 22 40


Output

0


Input

5 1000000000
6 6 2 6 2


Output

166666667

Solution (in JAVA):

Solution:
import java.io.InputStream;
import java.io.PrintWriter;
import java.lang.reflect.Array;
import java.math.BigDecimal;
import java.math.MathContext;
import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Comparator;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Iterator;
import java.util.List;
import java.util.Map;
import java.util.Map.Entry;
import java.util.PriorityQueue;
import java.util.Queue;
import java.util.Scanner;
import java.util.Set;
import java.util.Stack;
import java.util.TreeMap;
import java.util.TreeSet;

import static java.util.Comparator.*;

public class Main {
    public static void main(String[] args) {
        PrintWriter out = new PrintWriter(System.out);
        Solver solver = new Solver(System.in, out);
        solver.solve();
        out.close();
    }
}
class Solver {
	Scanner sc;
	PrintWriter out;
    public Solver(InputStream in, PrintWriter out) {
    	sc = new Scanner(in);
    	this.out = out;
    }
    // ==================================================================
    // 最大公約数の取得		
    long gcd(long a, long b) {		
    	if (b == 0)		return a;
    	return gcd(b, a % b);	
    }
    // 最小公倍数の取得			
    long lcm(long a, long b) {			
    	return a / gcd(a, b)  * b;		
    }
    int f(long x) {
    	int ans = 0;
    	while(x % 2 == 0) {
    		x /= 2;
    		ans++;
    	}
    	return ans;
    }
	public void solve() {
		int N = ni();
		long M = nl();
		long[] A = ndl(N);
		for (int i = 0; i < N; i++) {
			if(A[i] % 2 != 0) {
				out.println(0);
				return;
			}
			A[i] /= 2;
		}
		int t = f(A[0]);
		for (int i = 0; i < N; i++) {
			if(f(A[i]) != t) {
				out.println(0);
				return;
			}
			A[i] >>= t;
		}
		M >>= t;
		long lcm = A[0];
		for (int i = 1; i < N; i++) {
			lcm = lcm(lcm, A[i]);
			if(lcm > M) {
				out.println(0);
				return;
			}
//			out.println("lcm = " + lcm);
		}
		out.println(M < lcm ? 0 : 1 + (M-lcm)/(lcm*2));
	}
    static class PP{
    	public int key, val;
    	public PP(int key, int val) {
    		this.key = key;
    		this.val = val;
    	}
		public int getKey() {
			return key;
		}
		public void setKey(int key) {
			this.key = key;
		}
		public int getVal() {
			return val;
		}
		public void setVal(int val) {
			this.val = val;
		}
    }

    // ------------------------------------------
    // 入力
    // ------------------------------------------
    public int ni() {
    	return sc.nextInt();
    }
    public long nl() {
    	return sc.nextLong();
    }
    public String ns() {
    	return sc.next();
    }
    public char[] nc() {
    	return sc.next().toCharArray();
    }
    public int[] ndi(int N) {
    	int[] ans = new int[N];
    	for (int i = 0; i < N; i++) {
			ans[i] = ni();
		}
    	return ans;
    }
    public long[] ndl(int N) {
    	long[] ans = new long[N];
    	for (int i = 0; i < N; i++) {
			ans[i] = nl();
		}
    	return ans;
    }
    public String[] nds(int N) {
    	String[] ans = new String[N];
    	for (int i = 0; i < N; i++) {
			ans[i] = ns();
		}
    	return ans;
    }
    public char[][] ndc(int N) {
    	char[][] ans = new char[N][];
    	for (int i = 0; i < N; i++) {
			ans[i] = nc();
		}
    	return ans;
    }
    public int[][] nddi(int N, int S) {
    	int[][] ans = new int[N][S];
    	for (int i = 0; i < N; i++) {
			for (int j = 0; j < S; j++) {
				ans[i][j] = ni();
			}
		}
    	return ans;
    }
}