// Success consists of going from failure to failure without loss of enthusiasm
#include <bits/stdc++.h>
using namespace std;
#define nl '\n'
#define pb push_back
#define sz(x) int(x.size())
template<class T> using V = vector<T>;
using vi = V<int>;
int main() {
cin.tie(0)->sync_with_stdio(0);
int N; cin >> N;
vi D; unordered_map<int, int> I;
for(int i = 1; i * i <= N; i++) {
if (N % i == 0) {
D.pb(i);
if (i*i != N) D.pb(N / i);
}
}
int M = sz(D);
sort(begin(D), end(D)); for(int i = 0; i < M; i++) I[D[i]] = i;
V<vi> ans(M);
function<void(int)> dnc = [&](int x) {
int i = I[x];
if (sz(ans[i])) return;
ans[i] = {x - 1};
for(int y = 1; y * y <= x; y++) {
if (x % y == 0) {
dnc(y); dnc(x / y);
int ai = I[y], bi = I[x / y];
// cout << "MERGE " << x << endl;
// cout << y << " " << sz(ans[ai]) << endl;
// cout << x / y << " " << sz(ans[bi]) << endl;
cout << endl;
for(auto& a : ans[ai]) for(auto& b : ans[bi]) ans[i].pb(a + b);
}
}
sort(begin(ans[i]), end(ans[i]));
ans[i].erase(unique(begin(ans[i]), end(ans[i])), end(ans[i]));
};
dnc(N);
cout << sz(ans.back()) << nl;
for(auto& x : ans.back()) cout << x << " ";
cout << nl;
return 0;
}
