#include <bits/stdc++.h>
using namespace std;
long long squaresInSquare(long long m) {
  return (m * (m + 1)) / 2 * (2 * m + 1) / 3;
}
long long squaresInMN(long long m, long long n) {
  long long q = m * (m + 1) / 2;
  double q2 = (2 * m + 1) / 3.0;
  return q * q2 + (n - m) * q;
}
long long computeN(long long m, long long x) {
  long long q = m * (m + 1) / 2;
  long long q3 = q * (2 * m + 1) / 3;
  if ((x - q3) % q != 0) return -1;
  long long o = (x - q3) / q;
  return m + o;
}
struct MN {
  long long m;
  long long n;
  MN(long long a, long long b) : m(a), n(b) {}
};
void doProblem(long long x) {
  if (x == 1) {
    cout << "1" << endl << "1 1" << endl;
    return;
  }
  vector<MN> ans;
  ans.push_back(MN(1, x));
  for (long long m = 2; m < x; ++m) {
    if (squaresInSquare(m) > x) break;
    long long n = computeN(m, x);
    if (n == -1) continue;
    if (n < m) {
      cerr << "m, n: " << m << ", " << n << endl;
      break;
    }
    ans.push_back(MN(m, n));
  }
  bool backIsSquare = (ans.back().m == ans.back().n);
  long long cnt = ans.size() * 2;
  if (backIsSquare) cnt--;
  cout << cnt << endl;
  for (long long i = 0; i < ans.size(); ++i) {
    cout << ans[i].m << " " << ans[i].n << endl;
  }
  long long oStart = ans.size() - 1 - (backIsSquare ? 1 : 0);
  for (long long i = oStart; i >= 0; --i) {
    cout << ans[i].n << " " << ans[i].m << endl;
  }
}
int main(int, char **) {
  long long x;
  cin >> x;
  doProblem(x);
  return 0;
}

### Prompt

Please formulate a CPP solution to 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;
long long squaresInSquare(long long m) {
  return (m * (m + 1)) / 2 * (2 * m + 1) / 3;
}
long long squaresInMN(long long m, long long n) {
  long long q = m * (m + 1) / 2;
  double q2 = (2 * m + 1) / 3.0;
  return q * q2 + (n - m) * q;
}
long long computeN(long long m, long long x) {
  long long q = m * (m + 1) / 2;
  long long q3 = q * (2 * m + 1) / 3;
  if ((x - q3) % q != 0) return -1;
  long long o = (x - q3) / q;
  return m + o;
}
struct MN {
  long long m;
  long long n;
  MN(long long a, long long b) : m(a), n(b) {}
};
void doProblem(long long x) {
  if (x == 1) {
    cout << "1" << endl << "1 1" << endl;
    return;
  }
  vector<MN> ans;
  ans.push_back(MN(1, x));
  for (long long m = 2; m < x; ++m) {
    if (squaresInSquare(m) > x) break;
    long long n = computeN(m, x);
    if (n == -1) continue;
    if (n < m) {
      cerr << "m, n: " << m << ", " << n << endl;
      break;
    }
    ans.push_back(MN(m, n));
  }
  bool backIsSquare = (ans.back().m == ans.back().n);
  long long cnt = ans.size() * 2;
  if (backIsSquare) cnt--;
  cout << cnt << endl;
  for (long long i = 0; i < ans.size(); ++i) {
    cout << ans[i].m << " " << ans[i].n << endl;
  }
  long long oStart = ans.size() - 1 - (backIsSquare ? 1 : 0);
  for (long long i = oStart; i >= 0; --i) {
    cout << ans[i].n << " " << ans[i].m << endl;
  }
}
int main(int, char **) {
  long long x;
  cin >> x;
  doProblem(x);
  return 0;
}

```