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.InputStream;
import java.io.InputStreamReader;
import java.math.BigInteger;
import java.util.Arrays;
import java.util.Random;
import java.util.StringTokenizer;
import java.util.TreeSet;

public class C {

	static int count(long x) {
		int ret = 0;
		while (x > 0) {
			x /= 5;
			ret += x;
		}
		return ret;
	}

	public static void main(String[] args) throws IOException {
		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();

		long lo = 0;
		long hi = (long) (1e9);
		long start = -1;
		while (lo < hi) {
			long mid = (hi + lo) / 2;
			int zeros = count(mid);
			if (zeros >= N) {
				start = mid;
				hi = mid;
			} else {
				lo = mid + 1;
			}
		}
		StringBuilder ans = new StringBuilder();
		int c = 0;
		if (start != -1) {
			while (count(start) == N) {
				c++;
				ans.append((start++) + " ");
			}
		}
		System.out.println(c);
		System.out.println(ans);

	}

	static class Scanner {
		StringTokenizer st;
		BufferedReader br;

		public Scanner(InputStream s) {
			br = new BufferedReader(new InputStreamReader(s));
		}

		public String next() throws IOException {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}

		public int nextInt() throws IOException {
			return Integer.parseInt(next());
		}

		public long nextLong() throws IOException {
			return Long.parseLong(next());
		}

		public String nextLine() throws IOException {
			return br.readLine();
		}

		public double nextDouble() throws IOException {
			return Double.parseDouble(next());
		}

		public boolean ready() throws IOException {
			return br.ready();
		}

		public int[] nextIntArray(int n) throws IOException {
			int[] a = new int[n];
			for (int i = 0; i < n; i++)
				a[i] = nextInt();
			return a;
		}

		public int[] nextIntArray1(int n) throws IOException {
			int[] a = new int[n + 1];
			for (int i = 1; i <= n; i++)
				a[i] = nextInt();
			return a;
		}

		public int[] nextIntArraySorted(int n) throws IOException {
			int[] a = nextIntArray(n);
			Random r = new Random();
			for (int i = 0; i < n; i++) {
				int j = i + r.nextInt(n - i);
				int t = a[i];
				a[i] = a[j];
				a[j] = t;
			}
			Arrays.sort(a);
			return a;
		}

		public long[] nextLongArray(int n) throws IOException {
			long[] a = new long[n];
			for (int i = 0; i < n; i++)
				a[i] = nextLong();
			return a;
		}

		public long[] nextLongArray1(int n) throws IOException {
			long[] a = new long[n + 1];
			for (int i = 1; i <= n; i++)
				a[i] = nextLong();
			return a;
		}

		public long[] nextLongArraySorted(int n) throws IOException {
			long[] a = nextLongArray(n);
			Random r = new Random();
			for (int i = 0; i < n; i++) {
				int j = i + r.nextInt(n - i);
				long t = a[i];
				a[i] = a[j];
				a[j] = t;
			}
			Arrays.sort(a);
			return a;
		}
	}
}