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>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <typename T>
bool mmax(T &m, const T q) {
  if (m < q) {
    m = q;
    return true;
  } else
    return false;
}
template <typename T>
bool mmin(T &m, const T q) {
  if (m > q) {
    m = q;
    return true;
  } else
    return false;
}
template <typename T>
T add(T &a, T b) {
  return (a += b) %= 1000000007;
}
template <typename T>
T sub(T &a, T b) {
  (a -= b);
  if (a < 0) a += 1000000007;
}
long long power(long long a, long long b, long long m) {
  long long res = 1;
  while (b) {
    if (b & 1) {
      res = res * a;
      res %= m;
    }
    a = a * a;
    a %= m;
    b = b / 2;
  }
  return res;
}
long long inverse(long long first) {
  return power(first, 1000000007 - 2, 1000000007);
}
long long const N = 2e6 + 55;
long long fact[N], ifact[N];
void precompute() {
  fact[0] = 1;
  for (long long i = 1; i < N; i++) {
    fact[i] = fact[i - 1] * i;
    fact[i] %= 1000000007;
  }
  ifact[N - 1] = inverse(fact[N - 1]);
  for (long long i = N - 2; i >= 0; i--) {
    ifact[i] = (i + 1) * ifact[i + 1];
    ifact[i] %= 1000000007;
  }
}
set<pair<long long, long long> > ans;
set<pair<long long, long long> >::iterator it;
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long need;
  cin >> need;
  need *= 6;
  for (long long i = 1; i <= 2200000; i++) {
    long long rem = need + i * i * i - i;
    long long D2 = i * i + i;
    D2 *= 3;
    if (rem % D2) continue;
    long long val2 = rem / D2;
    if (val2 < i) continue;
    ans.insert(make_pair(i, val2));
    ans.insert(make_pair(val2, i));
  }
  cout << ans.size() << endl;
  for (it = ans.begin(); it != ans.end(); it++) {
    cout << (*it).first << " " << (*it).second << endl;
  }
  return 0;
}