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.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.util.ArrayList;

public class B {
	
	static int numZeros(int num) { 
		
		int counter = 0;
		
		while(num % 5 == 0) {
			num /= 5;
			counter++;
		}
		
		return counter;
	}
	public static void main(String [] args) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
		
		int M = Integer.parseInt(br.readLine());
		
		
		ArrayList<Integer> sol = new ArrayList<Integer>();
		
		int traveler = 5;
		int numZeros = 0;
		while(true) { 
			
			if(traveler % 5 == 0) { 
				int numFives = numZeros(traveler);
				numZeros += numFives;
			}
			
			if(numZeros > M) break;
			
			if(numZeros == M) sol.add(traveler);
			
			traveler++;
		}
		bw.write("" + sol.size());
		bw.newLine();
		for(int i=0; i<sol.size(); ++i) {
			
			if(i == sol.size() -1) { 
				bw.write("" + sol.get(i) );
				bw.newLine();
			} else { 
				bw.write(sol.get(i) + " ");
			}
		}
	
		bw.close();
	}
}