Codeforces Round #465 (Div. 2) A. Fafa and his Company 水题

A. Fafa and his Company

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Fafa owns a company that works on huge projects. There are n employees in Fafa's company. Whenever the company has a new project to start working on, Fafa has to divide the tasks of this project among all the employees.

Fafa finds doing this every time is very tiring for him. So, he decided to choose the best l employees in his company as team leaders. Whenever there is a new project, Fafa will divide the tasks among only the team leaders and each team leader will be responsible of some positive number of employees to give them the tasks. To make this process fair for the team leaders, each one of them should be responsible for the same number of employees. Moreover, every employee, who is not a team leader, has to be under the responsibility of exactly one team leader, and no team leader is responsible for another team leader.

Given the number of employees n, find in how many ways Fafa could choose the number of team leaders l in such a way that it is possible to divide employees between them evenly.

Input

The input consists of a single line containing a positive integer n (2 ≤ n ≤ 105) — the number of employees in Fafa's company.

Output

Print a single integer representing the answer to the problem.

Examples

input

Copy

2

output

1

input

Copy

10

output

3

Note

In the second sample Fafa has 3 ways:

• choose only 1 employee as a team leader with 9 employees under his responsibility.
• choose 2 employees as team leaders with 4 employees under the responsibility of each of them.

• choose 5 employees as team leaders with 1 employee under the responsibility of each of them.

题意：工厂给定一种管理方式，使得每个领导的手下个数均相等且领导互相平级，管理无交叉。给定工人的个数，求有多少种分配方案。

将工人总数减去领导数，再看领导数是否可被剩余人数整除即可。

代码如下：

#include 
using namespace std;


int main()
{
    int n;
    while (~scanf("%d",&n)){
        int ans=0;
        for (int i=1;i<=n/2;i++){
            int rest=n-i;
            if (rest%i==0)
                ans++;
        }
        printf("%d\n",ans);
    }
    return 0;
}