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;
int a[6000005];
vector<long long> primes;
vector<pair<long long, long long> > v;
void sieve() {
  int i, j;
  a[0] = 1;
  a[1] = 1;
  primes.push_back(2);
  for (i = 4; i < 6000005; i += 2) a[i] = 1;
  for (i = 3; i * i < 6000005; i += 2) {
    if (a[i] == 0) {
      primes.push_back(i);
      for (j = i + i; j < 6000005; j += i) a[j] = 1;
    }
  }
  for (; i < 6000005; i++)
    if (a[i] == 0) primes.push_back(i);
}
int main() {
  long long x, y, t1, m, n, i, j, t2;
  scanf("%lld", &x);
  sieve();
  t1 = x;
  for (i = 0; i < primes.size(); i++) {
    while (t1 % primes[i] == 0) {
      t1 /= primes[i];
    }
  }
  y = x / t1;
  y *= 6;
  for (i = 1; i < 6000005; i++) {
    if ((y % (i * (i + 1)) == 0)) {
      t1 = ((x * 6) / (i * (i + 1))) + (i - 1);
      if (t1 % 3 != 0) continue;
      t1 /= 3;
      if (i > t1) continue;
      v.push_back(make_pair(i, t1));
      if (i != t1) v.push_back(make_pair(t1, i));
    }
  }
  sort(v.begin(), v.end());
  printf("%d\n", (int)v.size());
  for (i = 0; i < v.size(); i++) printf("%lld %lld\n", v[i].first, v[i].second);
  return 0;
}