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;
const int maxn = 2000100;
long long sum[maxn];
long long sum2[maxn];
struct node {
  long long l;
  long long r;
};
vector<node> ans;
void init() {
  ans.clear();
  sum[0] = 0ll;
  sum2[0] = 0ll;
  for (long long i = 1; i < maxn; i++) sum[i] = sum[i - 1] + i;
  for (long long i = 1; i < maxn; i++) sum2[i] = sum2[i - 1] + i * i;
  long long ans = 0;
}
int main(int argc, char *argv[]) {
  init();
  long long x;
  cin >> x;
  for (long long i = 1; i <= 2e6; i++) {
    long long te = sum[i];
    long long tmp = sum2[i];
    if (tmp > x) break;
    if ((x - tmp) % te != 0ll)
      continue;
    else {
      long long as = (x - tmp) / te;
      node tt;
      tt.l = i;
      tt.r = as + i;
      if (tt.r < tt.l) break;
      ans.push_back(tt);
    }
  }
  long long cnt = 2 * ans.size();
  if (ans[ans.size() - 1].l == ans[ans.size() - 1].r) cnt--;
  printf("%d\n", cnt);
  for (int i = 0; i < ans.size(); i++) {
    printf("%I64d %I64d\n", ans[i].l, ans[i].r);
  }
  for (int i = ans.size() - 1; i >= 0; i--) {
    if (ans[i].l == ans[i].r) continue;
    printf("%I64d %I64d\n", ans[i].r, ans[i].l);
  }
  return 0;
}