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.

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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>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
const int mod2 = 998244353;
void fact_init(int n);
long long exp(long long taban, long long us, long long md);
long long ebob(long long a, long long b);
long long ekok(long long a, long long b);
long long komb(long long a, long long b);
vector<long long> fact;
vector<long long> inv_fact;
void fact_init(int n) {
  fact.resize(n + 5);
  inv_fact.resize(n + 5);
  fact[0] = inv_fact[0] = 1;
  for (int i = 1; i <= n; i++) {
    fact[i] = (fact[i - 1] * i) % mod;
    inv_fact[i] = exp(fact[i], mod - 2, mod);
  }
}
long long exp(long long taban, long long us, long long md) {
  long long carpan = taban % md;
  if (carpan == 0) return 0;
  long long temp = us;
  long long res = 1;
  while (temp) {
    if (temp % 2) res = (res * carpan) % md;
    temp /= 2;
    carpan = (carpan * carpan) % md;
  }
  return res;
}
long long ebob(long long a, long long b) {
  if (!a) return b;
  return ebob(b % a, a);
}
long long ekok(long long a, long long b) { return (a * b) / ebob(a, b); }
long long komb(long long a, long long b) {
  if (a < b) return 0;
  return fact[a] * (inv_fact[a - b] * inv_fact[b] % mod) % mod;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  vector<pair<long long, long long>> vec;
  long long m;
  cin >> m;
  m *= 6;
  bool check = false;
  for (long long a = 1; a * a * a <= m; a++) {
    long long ilk = a;
    long long iki = a + 1;
    long long uc = ilk + iki;
    long long tmp = m;
    if (tmp % ilk) continue;
    tmp /= ilk;
    if (tmp % iki) continue;
    tmp /= iki;
    tmp -= uc;
    if (tmp % 3 || tmp < 0) continue;
    tmp /= 3;
    vec.push_back({a, a + tmp});
    if (tmp == 0) check = true;
  }
  cout << 2 * vec.size() - check << "\n";
  for (auto x : vec) cout << x.first << " " << x.second << "\n";
  reverse(vec.begin(), vec.end());
  for (auto x : vec)
    if (x.first != x.second) cout << x.second << " " << x.first << "\n";
}