#include <bits/stdc++.h>
using namespace std;
long long n, a[20000005];
int shit = 0;
vector<pair<long long, long long> > v;
map<long long, long long> m;
void sq() {
  a[0] = 0;
  a[1] = 1;
  for (long long i = 2; i < 2000005; ++i) {
    a[i] = a[i - 1] + i * i;
  }
}
pair<long long, bool> solve2(long long now) {
  long long nn = n - a[now - 1] + now * now * (now - 1) / 2;
  if (nn < 0 || nn < now * now - (now * (now - 1) / 2))
    return make_pair(-1ll, 0);
  long long ans = (nn) / (now * now - (now * (now - 1) / 2));
  bool ok = 0;
  if ((nn) % (now * now - (now * (now - 1)) / 2) == 0) {
    ok = 1;
  }
  return make_pair(ans, ok);
}
void solve() {
  long long now = 2;
  v.push_back(make_pair(1, n));
  while (now < v.back().second) {
    pair<long long, bool> ans = solve2(now);
    if (ans.first < now && ans.first > 0) break;
    if (ans.first > 0 && ans.second) {
      v.push_back(make_pair(now, ans.first));
      if (ans.first == now) shit = 1;
    }
    now++;
  }
}
int main() {
  sq();
  while (cin >> n) {
    if (n == 1) {
      cout << 1 << endl << "1 1\n";
      continue;
    }
    v.clear();
    shit = 0;
    solve();
    cout << 2 * v.size() - shit << endl;
    for (int i = 0; i < v.size(); ++i) {
      cout << v[i].first << " " << v[i].second << endl;
    }
    for (int i = v.size() - 1; i > -1; --i) {
      if (v[i].first != v[i].second)
        cout << v[i].second << " " << v[i].first << endl;
    }
  }
}

### 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>
using namespace std;
long long n, a[20000005];
int shit = 0;
vector<pair<long long, long long> > v;
map<long long, long long> m;
void sq() {
  a[0] = 0;
  a[1] = 1;
  for (long long i = 2; i < 2000005; ++i) {
    a[i] = a[i - 1] + i * i;
  }
}
pair<long long, bool> solve2(long long now) {
  long long nn = n - a[now - 1] + now * now * (now - 1) / 2;
  if (nn < 0 || nn < now * now - (now * (now - 1) / 2))
    return make_pair(-1ll, 0);
  long long ans = (nn) / (now * now - (now * (now - 1) / 2));
  bool ok = 0;
  if ((nn) % (now * now - (now * (now - 1)) / 2) == 0) {
    ok = 1;
  }
  return make_pair(ans, ok);
}
void solve() {
  long long now = 2;
  v.push_back(make_pair(1, n));
  while (now < v.back().second) {
    pair<long long, bool> ans = solve2(now);
    if (ans.first < now && ans.first > 0) break;
    if (ans.first > 0 && ans.second) {
      v.push_back(make_pair(now, ans.first));
      if (ans.first == now) shit = 1;
    }
    now++;
  }
}
int main() {
  sq();
  while (cin >> n) {
    if (n == 1) {
      cout << 1 << endl << "1 1\n";
      continue;
    }
    v.clear();
    shit = 0;
    solve();
    cout << 2 * v.size() - shit << endl;
    for (int i = 0; i < v.size(); ++i) {
      cout << v[i].first << " " << v[i].second << endl;
    }
    for (int i = v.size() - 1; i > -1; --i) {
      if (v[i].first != v[i].second)
        cout << v[i].second << " " << v[i].first << endl;
    }
  }
}

```