Problem: Mr. Santa asks all the great programmers of the world to solve a trivial problem. He gives them an integer m and asks for the number of positive integers n, such that the factorial of n ends with exactly m zeroes. Are you among those great programmers who can solve this problem?

Input

The only line of input contains an integer m (1 ≤ m ≤ 100 000) — the required number of trailing zeroes in factorial.

Output

First print k — the number of values of n such that the factorial of n ends with m zeroes. Then print these k integers in increasing order.

Examples

Input

1


Output

5
5 6 7 8 9 

Input

5


Output

0

Note

The factorial of n is equal to the product of all integers from 1 to n inclusive, that is n! = 1·2·3·...·n.

In the first sample, 5! = 120, 6! = 720, 7! = 5040, 8! = 40320 and 9! = 362880.

Solution (in JAVA):

Solution:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

public class ATrivialProblem {
	public static BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
	public static StringTokenizer st;
	
	public static void main(String[] args) throws IOException {
		int m = nextInt();
		int x = 0;
		int sum = 0;
		
		while (sum < m) {
			x += 5;
			int temp = x;
			while (temp % 5 == 0) {
				sum++;
				temp /= 5;
			}
		}
		
		if (sum == m) {
			System.out.println(5);
			for (int i = 0; i < 5; i++)
				System.out.print((x+i) + " ");
			System.out.println();
		}
		else
			System.out.println(0);
	}
	
	public static String nextLine() throws IOException {
		return f.readLine();
	}
	
	public static String nextString() throws IOException {
		while (st == null || !st.hasMoreTokens())
			st = new StringTokenizer(f.readLine());
		return st.nextToken();
	}
	
	public static int nextInt() throws IOException {
		return Integer.parseInt(nextString());
	}
	
	public static long nextLong() throws IOException {
		return Long.parseLong(nextString());
	}
	
	public static int[] intArray(int n) throws IOException {
		int[] a = new int[n];
		for (int i = 0; i < n; i++)
			a[i] = nextInt();
		return a;
	}
	
	public static int[][] intArray(int n, int m) throws IOException {
		int[][] a = new int[n][m];
		for (int i = 0; i < n; i++)
			for (int j = 0; j < m; j++)
				a[i][j] = nextInt();
		return a;
	}
	
	public static long[] longArray(int n) throws IOException {
		long[] a = new long[n];
		for (int i = 0; i < n; i++)
			a[i] = nextLong();
		return a;
	}
}