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>
#pragma comment(linker, "/STACK:256000000")
using namespace std;
typedef long long ll;
void my_return(ll code) { exit(code); }
ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
ll my_exp(ll a, ll b, ll MOD = numeric_limits<ll>::max()) {
  a %= MOD;
  if (b == 0) {
    return 1;
  } else {
    return ((b % 2 == 1 ? a : 1) * my_exp(a * a, b / 2, MOD)) % MOD;
  }
}
bool is_prime(ll n) {
  if (n != 2 && n % 2 == 0) {
    return false;
  } else {
    for (int i = 3; (i * i) <= n; i += 2) {
      if (n % i == 0) return false;
    }
    return true;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  ll x;
  cin >> x;
  vector<pair<ll, ll> > ans;
  for (ll i = 1; i < 2000000; ++i) {
    if ((x + ((i * (i + 1) * (i - 1)) / 6)) % (i * (i + 1) / 2) == 0) {
      ll n = (x + ((i * (i + 1) * (i - 1)) / 6)) / (i * (i + 1) / 2);
      if (n > 0 && n >= i) {
        ans.push_back(make_pair(i, n));
      }
    }
  }
  ll s = 2 * ll(ans.size());
  for (decltype(ans.begin()) it = ans.begin(); it != ans.end(); it++) {
    if (it->first == it->second) {
      --s;
      break;
    }
  }
  cout << s << endl;
  for (decltype(ans.begin()) it = ans.begin(); it != ans.end(); it++) {
    cout << it->first << " " << it->second << endl;
  }
  for (decltype(ans.rbegin()) it = ans.rbegin(); it != ans.rend(); it++) {
    if (it->first != it->second) {
      cout << it->second << " " << it->first << endl;
    }
  }
  my_return(0);
}