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;
void solve(int);
int main() {
  solve(0);
  return 0;
}
long long dvd[2000000], off[2000000];
void solve(int tc) {
  memset(dvd, 0, sizeof(dvd));
  for (int x = 1, _c = 2000000; x < _c; ++x) dvd[x] = dvd[x - 1] + x;
  for (int x = 1, _c = 2000000; x < _c; ++x) off[x] = off[x - 1] + dvd[x - 1];
  long long i;
  scanf("%I64d", &i);
  queue<pair<long long, long long> > q;
  stack<pair<long long, long long> > s;
  for (int x = 1, _c = 2000000; x < _c; ++x) {
    long long y = (i + off[x]) / dvd[x];
    if (y < x) break;
    if ((i + off[x]) % dvd[x]) continue;
    if (y > x) s.push(make_pair(y, (long long)x));
    q.push(make_pair((long long)x, y));
  }
  while (!s.empty()) {
    q.push(s.top());
    s.pop();
  }
  printf("%d\n", q.size());
  while (!q.empty()) {
    pair<long long, long long> tmp = q.front();
    q.pop();
    printf("%I64d %I64d\n", tmp.first, tmp.second);
  }
}