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: Time Limit Exceeded)

#include <iostream>
#include <string>
#include <vector>
#include <algorithm>
using namespace std;

typedef long long ll;
typedef pair<ll,ll> P;

vector<P> prifac(ll n){
	vector<P> ret;
	for(int i = 2 ; i*i <= n ; i ++){
		if(n%i == 0){
			int cnt = 0;
			while(n%i == 0){
				cnt ++;
				n /= i;
			}
			ret.push_back(P(i,cnt));
		}
	}
	if(n > 1)ret.push_back(P(n,1));
	return ret;
}

vector<ll> divisors(ll n){
	vector<P> vec = prifac(n);
	vector<ll> ret;
	ret.push_back(1);
	for(int i = 0 ; i < vec.size() ; i ++){
		P p = vec[i];
		int m = ret.size();
		for(int j = 0 ; j < m ; j ++){
			ll t = ret[j];
			for(int k = 0 ; k < p.second ; k ++){
				t *= p.first;
				ret.push_back(t);
			}
		}
	}
	return ret;
}

int main(){
	ll n;
	cin >> n;
	
	vector<ll> vec = divisors(n);
	ll ret = 0;
	for(int i = 0 ; i < vec.size() ; i ++){
		if(n/vec[i]+2 <= vec[i])ret += vec[i]-1;
	}
	cout << ret << endl;
}