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: Accepted)

#include <bits/stdc++.h>
using namespace std;

//#define int long long
using ll = long long;
const int INF = 1<<29;
const int MOD = (int)1e9 + 7;
const double EPS = 1e-9;
const int dx[4]={1,0,-1,0};
const int dy[4]={0,1,0,-1};
#ifdef LOCAL_ENV
	#define debug(var) std::cout<<#var" = "<<var<<std::endl
#else
	#define debug(var)
#endif
#define p(var)  std::cout<<var<<std::endl
#define PI (acos(-1))
#define rep(i, n)   for(int i=0, i##_length=(n); i< i##_length; ++i)
#define repeq(i, n) for(int i=1, i##_length=(n); i<=i##_length; ++i)
#define all(a) (a).begin(), (a).end()
#define pb push_back
inline bool isnatural(double a) { return a>=0 && ceil(a)==floor(a); }
template<typename T> inline T gcd(T a, T b) { return b ? gcd(b,a%b) : a; }
template<typename T> inline T lcm(T a, T b) { return a / gcd(a,  b) * b; }
template<typename T1, typename T2>inline bool chmax(T1 &a, T2 b) { return a < b && (a = b, true); }
template<typename T1, typename T2>inline bool chmin(T1 &a, T2 b) { return a > b && (a = b, true); }
template<typename T1, typename T2> ostream& operator<<(ostream& s, const pair<T1, T2>& p) {return s << "(" << p.first << ", " << p.second << ")";}
template<typename T> ostream& operator<<(ostream& s, const vector<T>& v) {
	for (int i = 0, len = v.size(); i < len; ++i){
		s << v[i]; if (i < len - 1) s << "\t";
	}
	return s;
}
template<typename T> ostream& operator<<(ostream& s, const vector< vector<T> >& vv) {
	for (int i = 0, len = vv.size(); i < len; ++i){
		s << vv[i] << endl;
	}
	return s;
}
template<typename T1, typename T2> ostream& operator<<(ostream& s, const map<T1, T2>& m) {
	s << "{" << endl;
	for (typeof(m.begin()) itr = m.begin(); itr != m.end(); ++itr){
		s << "\t" << (*itr).first << " : " << (*itr).second << endl;
	}
	s << "}" << endl;
	return s;
}

/*-----8<-----8<-----*/

const ll N_MAX = 1000000000000;

ll N;


//nの約数を返す　ソートされていないので必要に応じてsort(all(yakusuu));してください
template<typename T>
vector<T> divisor(T n){
	vector<T> div;
	for(T i=1 ; i*i<=n ; ++i){
		if(n%i == 0){
			div.push_back(i);
			if(i!=1 && i*i!=n){
				div.push_back(n/i);
			}
		}
	}
	div.push_back(n);
	return div;
}


void solve() {
	ll ans = 0;

	vector<ll> div=divisor(N);
	for(ll d:div){
		if(d==1)continue;
		ll dd=d-1;
		if(N/dd==N%dd){
			ans+=dd;
		}
	}

	printf("%lld\n", ans);
}

void input() {
	scanf("%lld", &N);
}

signed main() {
	input();
	solve();
	return 0;
}