#include <bits/stdc++.h>
#pragma optimization_level 3
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math,O3")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx")
using namespace std;
using ll = long long;
using ld = long double;
const ll MOD = 1e+9 + 7;
const ll INF = LLONG_MAX;
const int N = (int)2e+5 + 8;
ld x;
vector<pair<ll, ll>> res;
void MAIN(int tc) {
  cin >> x;
  for (ld a = 1; a <= 1e7; a++) {
    ld num = (6 * x) + (a * a * a) - (a);
    ld den = (3 * a * a) + (3 * a);
    ld b = num / den;
    if (fmod(num, den) == 0 && a <= b) {
      res.push_back({a, b});
      res.push_back({b, a});
    }
  }
  sort((res).begin(), (res).end(), [&](pair<ll, ll> p1, pair<ll, ll> p2) {
    return p1.first == p2.first ? p1.first < p2.first : p1.second > p2.second;
  });
  res.resize(unique(res.begin(), res.end()) - res.begin());
  ;
  cout << res.size() << "\n";
  ;
  for (auto [a, b] : res) {
    cout << a << " " << b << "\n";
    ;
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cout << fixed;
  cout << setprecision(10);
  int test_cases = 1;
  for (int i = 1; i <= test_cases; i++) {
    MAIN(i);
  }
}

### Prompt

Please provide a cpp coded solution to the problem described below:

Spongebob is already tired trying to reason his weird actions and calculations, so he simply asked you to find all pairs of n and m, such that there are exactly x distinct squares in the table consisting of n rows and m columns. For example, in a 3 × 5 table there are 15 squares with side one, 8 squares with side two and 3 squares with side three. The total number of distinct squares in a 3 × 5 table is 15 + 8 + 3 = 26.

Input

The first line of the input contains a single integer x (1 ≤ x ≤ 1018) — the number of squares inside the tables Spongebob is interested in.

Output

First print a single integer k — the number of tables with exactly x distinct squares inside.

Then print k pairs of integers describing the tables. Print the pairs in the order of increasing n, and in case of equality — in the order of increasing m.

Examples

Input

26


Output

6
1 26
2 9
3 5
5 3
9 2
26 1


Input

2


Output

2
1 2
2 1


Input

8


Output

4
1 8
2 3
3 2
8 1

Note

In a 1 × 2 table there are 2 1 × 1 squares. So, 2 distinct squares in total.

<image>

In a 2 × 3 table there are 6 1 × 1 squares and 2 2 × 2 squares. That is equal to 8 squares in total.

<image>

### Response

```cpp
#include <bits/stdc++.h>
#pragma optimization_level 3
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math,O3")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx")
using namespace std;
using ll = long long;
using ld = long double;
const ll MOD = 1e+9 + 7;
const ll INF = LLONG_MAX;
const int N = (int)2e+5 + 8;
ld x;
vector<pair<ll, ll>> res;
void MAIN(int tc) {
  cin >> x;
  for (ld a = 1; a <= 1e7; a++) {
    ld num = (6 * x) + (a * a * a) - (a);
    ld den = (3 * a * a) + (3 * a);
    ld b = num / den;
    if (fmod(num, den) == 0 && a <= b) {
      res.push_back({a, b});
      res.push_back({b, a});
    }
  }
  sort((res).begin(), (res).end(), [&](pair<ll, ll> p1, pair<ll, ll> p2) {
    return p1.first == p2.first ? p1.first < p2.first : p1.second > p2.second;
  });
  res.resize(unique(res.begin(), res.end()) - res.begin());
  ;
  cout << res.size() << "\n";
  ;
  for (auto [a, b] : res) {
    cout << a << " " << b << "\n";
    ;
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cout << fixed;
  cout << setprecision(10);
  int test_cases = 1;
  for (int i = 1; i <= test_cases; i++) {
    MAIN(i);
  }
}

```