#include <bits/stdc++.h>
using namespace std;
vector<vector<long long> > matmul(vector<vector<long long> > &a,
                                  vector<vector<long long> > &b) {
  int i, j, k, n = a.size();
  vector<vector<long long> > ans(n, vector<long long>(n));
  for (i = 0; i < n; i++)
    for (j = 0; j < n; j++) {
      long long temp = 0;
      for (k = 0; k < n; k++)
        temp = (temp + a[i][k] * b[k][j]) % (1000000007LL);
      ans[i][j] = temp;
    }
  return ans;
}
vector<vector<long long> > matpwr(vector<vector<long long> > &a, long long p) {
  int n = a.size();
  vector<vector<long long> > ans(n, vector<long long>(n));
  for (int i = 0; i < n; i++) ans[i][i] = 1;
  while (p) {
    if (p & 1) ans = matmul(ans, a);
    a = matmul(a, a);
    p >>= 1;
  }
  return ans;
}
long long pwr(long long base, long long p, long long mod = (1000000007LL)) {
  long long ans = 1;
  while (p) {
    if (p & 1) ans = (ans * base) % mod;
    base = (base * base) % mod;
    p /= 2;
  }
  return ans;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
long long invMod(long long n, long long mod = (1000000007LL)) {
  return pwr(n, mod - 2);
}
vector<pair<long long, long long> > ans;
int main() {
  ios_base::sync_with_stdio(0);
  long long x;
  cin >> x;
  for (long long side = 1; (side * (side + 1) * (2 * side + 1) / 6) <= x;
       side++) {
    if (side * (side + 1) * (2 * side + 1) / 6 == x)
      ans.push_back(pair<long long, long long>(side, side));
    long long rhs = x + side * (side - 1) * (side) / 2 -
                    side * (side - 1) * (2 * side - 1) / 6;
    long long lhs = side * side - (side - 1) * (side) / 2;
    if (rhs % lhs == 0 && (rhs / lhs) > side) {
      ans.push_back(pair<long long, long long>(side, rhs / lhs));
      ans.push_back(pair<long long, long long>(rhs / lhs, side));
    }
  }
  sort(ans.begin(), ans.end());
  cout << ans.size() << endl;
  for (int i = 0; i < (int)ans.size(); i++)
    cout << ans[i].first << " " << ans[i].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;
vector<vector<long long> > matmul(vector<vector<long long> > &a,
                                  vector<vector<long long> > &b) {
  int i, j, k, n = a.size();
  vector<vector<long long> > ans(n, vector<long long>(n));
  for (i = 0; i < n; i++)
    for (j = 0; j < n; j++) {
      long long temp = 0;
      for (k = 0; k < n; k++)
        temp = (temp + a[i][k] * b[k][j]) % (1000000007LL);
      ans[i][j] = temp;
    }
  return ans;
}
vector<vector<long long> > matpwr(vector<vector<long long> > &a, long long p) {
  int n = a.size();
  vector<vector<long long> > ans(n, vector<long long>(n));
  for (int i = 0; i < n; i++) ans[i][i] = 1;
  while (p) {
    if (p & 1) ans = matmul(ans, a);
    a = matmul(a, a);
    p >>= 1;
  }
  return ans;
}
long long pwr(long long base, long long p, long long mod = (1000000007LL)) {
  long long ans = 1;
  while (p) {
    if (p & 1) ans = (ans * base) % mod;
    base = (base * base) % mod;
    p /= 2;
  }
  return ans;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
long long invMod(long long n, long long mod = (1000000007LL)) {
  return pwr(n, mod - 2);
}
vector<pair<long long, long long> > ans;
int main() {
  ios_base::sync_with_stdio(0);
  long long x;
  cin >> x;
  for (long long side = 1; (side * (side + 1) * (2 * side + 1) / 6) <= x;
       side++) {
    if (side * (side + 1) * (2 * side + 1) / 6 == x)
      ans.push_back(pair<long long, long long>(side, side));
    long long rhs = x + side * (side - 1) * (side) / 2 -
                    side * (side - 1) * (2 * side - 1) / 6;
    long long lhs = side * side - (side - 1) * (side) / 2;
    if (rhs % lhs == 0 && (rhs / lhs) > side) {
      ans.push_back(pair<long long, long long>(side, rhs / lhs));
      ans.push_back(pair<long long, long long>(rhs / lhs, side));
    }
  }
  sort(ans.begin(), ans.end());
  cout << ans.size() << endl;
  for (int i = 0; i < (int)ans.size(); i++)
    cout << ans[i].first << " " << ans[i].second << endl;
  return 0;
}

```