Manthan, Codefest 16 B 数学

B. A Trivial Problem

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

题意：统计哪些数阶乘的末尾有m位‘0’    输出个数 并按照升序输出这些数

题解：数学知识  涨姿势

        我们知道阶乘下 产生‘0’位只能是 2*5=10产生‘0’位

        那么‘0’位的个数取决于‘2’和‘5’的个数

        相比之下‘5’的个数更少  所以末尾‘0’的个数取决于阶乘中‘5’的数

        所以可以预处理出所有数含有因数‘5’的个数 也就代码中fun的作用

        vector存储为一个数x 对应的x！中5的个数也就是结果。

       

#include<bits/stdc++.h>
#define ll __int64
#define mod 1e9+7
#define PI acos(-1.0)
#define bug(x) printf("%%%%%%%%%%%%%",x);
#define inf 1e8
using namespace std;
#define ll __int64
int fun( int x)
{
    int flag=0;
    if(x%5==0)
    {
        while(x%5==0)
        {
            x/=5;
            flag++;
        }
    }
    return flag;
}
vector<int>ve[100005];
int m;
int main()
{
    scanf("%d",&m);
    int gg=0;
    for(int i=1;i<=500000;i++)
    {
        gg+=fun(i);
        if(gg==m)
        ve[gg].push_back(i);
    }
    printf("%d
",ve[m].size());
    if(ve[m].size())
    cout<<ve[m][0];
    for(int i=1;i<ve[m].size();i++)
        cout<<" "<<ve[m][i];
    cout<<endl;
    return 0;
}