CF633B.A Trivial 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?

圣诞老人先生请全世界的优秀程序员解决一个简单的问题。他给出一个整数 $m$，要求找出满足 $n!$ 的十进制表示恰好以 $m$ 个零结尾的正整数 $n$ 的个数。你是否是能够解决这一问题的优秀程序员之一？

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

输入仅包含一个整数 $m$（$1 \leq m \leq 100\,000$）——阶乘结果中所需的末尾零的个数。

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.

首先输出 k —— 即满足 n 的阶乘末尾恰好有 m 个零的 n 的个数。然后按升序输出这 k 个整数。

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.

$n$ 的阶乘等于从 $1$ 到 $n$（含）所有整数的乘积，即 $n! = 1 \cdot 2 \cdot 3 \cdot \ldots \cdot n$。

在第一个样例中，$5! = 120$，$6! = 720$，$7! = 5040$，$8! = 40320$，且 $9! = 362880$。