Problem:
Score : 500 points

Problem Statement

Snuke received a positive integer N from Takahashi.
A positive integer m is called a favorite number when the following condition is satisfied:

The quotient and remainder of N divided by m are equal, that is, \lfloor \frac{N}{m} \rfloor = N \bmod m holds.

Find all favorite numbers and print the sum of those.

Constraints

All values in input are integers.

1 \leq N \leq 10^{12}

Input

Input is given from Standard Input in the following format:

N

Output

Print the answer.

Sample Input 1

8

Sample Output 1

10

There are two favorite numbers: 3 and 7. Print the sum of these, 10.

Sample Input 2

1000000000000

Sample Output 2

2499686339916

Watch out for overflow.

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
vector<long long> v;
struct hugeint{
	int len;
	long long num[200];
}a;
hugeint operator +(hugeint a,long long b){
	a.num[1]+=b;
	int i=1;
	while(a.num[i]>=10)
	{
		a.num[i+1]+=a.num[i]/10;
		a.num[i]%=10;
		i++;
	}
	a.len=max(a.len,i);
	return a;
}
void divi(long long n){
	for(long long i=1;i*i<=n;i++){
		if(n%i==0){
			v.push_back(i);
			v.push_back(n/i);
		}
	}
}
int main(){
	//freopen("input.txt","r",stdin);
	//freopen("output.txt","w",stdout);
	long long n;
	cin>>n;
	divi(n);
	a.len=0;
	for(int i=0;i<v.size();i++){
		long long m=n/v[i] -1;
		if(v[i]<m)
			a=a+m;
	}
	for(int i=a.len;i>=1;i--)
		cout<<a.num[i];
	cout<<endl;
	return 0;
}