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=;
     if(x%==)
     {
         while(x%==)
         {
             x/=;
             flag++;
         }
     }
     return flag;
 }
 vector<int>ve[];
 int m;
 int main()
 {
     scanf("%d",&m);
     int gg=;
     for(int i=;i<=;i++)
     {
         gg+=fun(i);
         if(gg==m)
         ve[gg].push_back(i);
     }
     printf("%d\n",ve[m].size());
     if(ve[m].size())
     cout<<ve[m][];
     for(int i=;i<ve[m].size();i++)
         cout<<" "<<ve[m][i];
     cout<<endl;
     return ;
 }