#include <bits/stdc++.h>
using namespace std;
const int inf = 0x20202020;
const long long linf = (((long long)1) << 60) - 1;
const long long mod = 1000000007;
const double eps = 1e-9;
const double pi = 3.1415926535897932384626;
const int DX[] = {1, 0, -1, 0}, DY[] = {0, 1, 0, -1};
long long powmod(long long a, long long b, long long mod) {
  long long res = 1;
  a %= mod;
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
  }
  return res;
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long inv_mod(long long a, long long mod) {
  return powmod(a, mod - 2, mod);
}
vector<pair<long long, long long> > V, V1;
int main() {
  std::ios::sync_with_stdio(false);
  long long X;
  cin >> X;
  long long sm = 0, sq = 0;
  for (long long N = 1; N * N <= X && N <= 10000000; N++) {
    sm = sm + N;
    sq = sq + (N * N);
    long long up = X + (N * sm) - sq + (2 * sm) - (N * N) - N;
    long long dn = (N * N) - sm + N;
    if (up % dn == 0) {
      long long M = up / dn;
      if (M >= N) {
        V1.push_back(make_pair(N, M));
        if (N != M) {
          V.push_back(make_pair(M, N));
        }
      }
    }
  }
  cout << ((int)(V).size()) + ((int)(V1).size()) << endl;
  for (int i = 0; i < ((int)(V1).size()); i++) {
    cout << V1[i].first << " " << V1[i].second << endl;
  }
  for (int i = ((int)(V).size()) - 1; i >= 0; i--) {
    cout << V[i].first << " " << V[i].second << endl;
  }
  return 0;
}

### Prompt

Develop a solution in Cpp 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;
const int inf = 0x20202020;
const long long linf = (((long long)1) << 60) - 1;
const long long mod = 1000000007;
const double eps = 1e-9;
const double pi = 3.1415926535897932384626;
const int DX[] = {1, 0, -1, 0}, DY[] = {0, 1, 0, -1};
long long powmod(long long a, long long b, long long mod) {
  long long res = 1;
  a %= mod;
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
  }
  return res;
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long inv_mod(long long a, long long mod) {
  return powmod(a, mod - 2, mod);
}
vector<pair<long long, long long> > V, V1;
int main() {
  std::ios::sync_with_stdio(false);
  long long X;
  cin >> X;
  long long sm = 0, sq = 0;
  for (long long N = 1; N * N <= X && N <= 10000000; N++) {
    sm = sm + N;
    sq = sq + (N * N);
    long long up = X + (N * sm) - sq + (2 * sm) - (N * N) - N;
    long long dn = (N * N) - sm + N;
    if (up % dn == 0) {
      long long M = up / dn;
      if (M >= N) {
        V1.push_back(make_pair(N, M));
        if (N != M) {
          V.push_back(make_pair(M, N));
        }
      }
    }
  }
  cout << ((int)(V).size()) + ((int)(V1).size()) << endl;
  for (int i = 0; i < ((int)(V1).size()); i++) {
    cout << V1[i].first << " " << V1[i].second << endl;
  }
  for (int i = ((int)(V).size()) - 1; i >= 0; i--) {
    cout << V[i].first << " " << V[i].second << endl;
  }
  return 0;
}

```