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;
using i64 = long long;
using Pll = pair<i64, i64>;
constexpr int MAX = 2000100;
constexpr int INF = 1000100;
Pll ans[MAX];
inline i64 Sum1(i64 n) { return n * (n + 1) / 2; }
inline i64 Sum2(i64 n) {
  i64 ret = n * (n + 1) / 2;
  if (!(ret % 3)) return ret / 3 * (2 * n + 1);
  return (2 * n + 1) / 3 * ret;
}
inline i64 Calc(i64 n, i64 d) { return Sum2(n) + d * Sum1(n); }
inline i64 Find(i64 n, i64 x) {
  i64 s1 = Sum1(n), s2 = Sum2(n);
  i64 ret = (x - s2) / s1;
  return ret * s1 + s2 == x ? ret : -1;
}
int Solve(i64 x) {
  int ret = 0;
  for (int i = 1; Sum2(i) <= x; ++i) {
    i64 cur = Find(i, x);
    if (cur != -1) {
      ans[ret++] = {(i64)i, cur + i};
      ans[ret++] = {cur + i, (i64)i};
    }
  }
  sort(ans, ans + ret);
  ret = unique(ans, ans + ret) - ans;
  return ret;
}
int main() {
  i64 x;
  while (cin >> x) {
    int cnt = Solve(x);
    printf("%d\n", cnt);
    for (int i = 0; i < cnt; ++i) {
      cout << ans[i].first << " " << ans[i].second << endl;
    }
  }
  return 0;
}