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.math.BigInteger;

/**
 * @author mohanad
 *
 */
public class P9_manthan_633B {

	/**
	 * @param args
	 * @throws IOException
	 * @throws NumberFormatException
	 */
	public static void main(String[] args) throws NumberFormatException, IOException {
		BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));

		int m = Integer.parseInt(bf.readLine());

		int l=1,h=600000,tmp ,ans=0;
		
		while (l<=h){
			tmp=l+(h-l)/2;
			
			int zers=numZeros(tmp);
			
			if (zers>m){
				h=tmp-1;
			}
			else if (zers<m){
				l=tmp+1;
			}
			else {
				ans=tmp;
				break;
			}
		}
		
		if (ans==0){
			System.out.println("0");
			return;
		}
		
		switch (ans%5) {
		case 1:
			--ans;
			break;
		case 2:
			ans-=2;
			break;
		case 3:
			ans-=3;
			break;
		case 4:
			ans-=4;
		default:
			break;
		}
		
		
		
		System.out.println(5);
		for (int i = 0 ; i< 5 ; ++i)
			System.out.print(ans+i+" ");
		
		// debug
		// System.out.println("\n"+fact(m).toString());

	}
	
	
	static int numZeros(int n){
		
		int ans=0, i =1 , curr=(int)Math.pow(5, 1);
		
		while (curr<=n){
			ans+=n/curr;
			++i;
			curr=(int)Math.pow(5, i);
		}
		
		return ans;
	}

	// debug
	static BigInteger fact(int n) {

		long ans = 1;
		BigInteger bg = new BigInteger("1");

		for (int i = 2; i <= n; ++i) {
			ans *= i;
			bg = bg.multiply(new BigInteger(i + ""));
		}

		return bg;
	}

}