#include <bits/stdc++.h>
using namespace std;
const int INF = 1 << 29;
inline int two(int n) { return 1 << n; }
long long extgcd(long long a, long long b, long long& x, long long& y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  } else {
    int g = extgcd(b, a % b, y, x);
    y -= a / b * x;
    return g;
  }
}
long long modpow(long long a, long long b, long long md) {
  long long res = 1;
  a %= md;
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % md;
    a = a * a % md;
  }
  return res;
}
struct custom_hash {
  static uint64_t splitmix64(uint64_t x) {
    x += 0x9e3779b97f4a7c15;
    x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
    x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
    return x ^ (x >> 31);
  }
  size_t operator()(uint64_t x) const {
    static const uint64_t FIXED_RANDOM =
        chrono::steady_clock::now().time_since_epoch().count();
    return splitmix64(x + FIXED_RANDOM);
  }
};
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ifstream in("input.txt");
  ofstream out("output.txt");
  long long x;
  cin >> x;
  long long xx = 6 * x, div;
  vector<pair<long long, long long> > res;
  for (long long l = 1; l * (l + 1) * (2 * l + 1) <= xx; l++) {
    if (xx % (l * (l + 1)) == 0) {
      div = (xx / (l * (l + 1)) + l - 1);
      if (div / 3 < l) break;
      if (div % 3 == 0) {
        res.push_back(make_pair(l, div / 3));
        if (3 * l != div) res.push_back(make_pair(div / 3, l));
      }
    }
  }
  sort(res.begin(), res.end());
  cout << res.size() << endl;
  for (pair<long long, long long> p : res)
    cout << p.first << " " << p.second << endl;
  return 0;
}

### Prompt

Create a solution in CPP for the following problem:

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;
const int INF = 1 << 29;
inline int two(int n) { return 1 << n; }
long long extgcd(long long a, long long b, long long& x, long long& y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  } else {
    int g = extgcd(b, a % b, y, x);
    y -= a / b * x;
    return g;
  }
}
long long modpow(long long a, long long b, long long md) {
  long long res = 1;
  a %= md;
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % md;
    a = a * a % md;
  }
  return res;
}
struct custom_hash {
  static uint64_t splitmix64(uint64_t x) {
    x += 0x9e3779b97f4a7c15;
    x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
    x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
    return x ^ (x >> 31);
  }
  size_t operator()(uint64_t x) const {
    static const uint64_t FIXED_RANDOM =
        chrono::steady_clock::now().time_since_epoch().count();
    return splitmix64(x + FIXED_RANDOM);
  }
};
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ifstream in("input.txt");
  ofstream out("output.txt");
  long long x;
  cin >> x;
  long long xx = 6 * x, div;
  vector<pair<long long, long long> > res;
  for (long long l = 1; l * (l + 1) * (2 * l + 1) <= xx; l++) {
    if (xx % (l * (l + 1)) == 0) {
      div = (xx / (l * (l + 1)) + l - 1);
      if (div / 3 < l) break;
      if (div % 3 == 0) {
        res.push_back(make_pair(l, div / 3));
        if (3 * l != div) res.push_back(make_pair(div / 3, l));
      }
    }
  }
  sort(res.begin(), res.end());
  cout << res.size() << endl;
  for (pair<long long, long long> p : res)
    cout << p.first << " " << p.second << endl;
  return 0;
}

```