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 N = 1e5 + 5;
const int MOD = 1e9 + 7;
const int INF = 0x3f3f3f3f;
pair<long long, long long> ans1[N], ans2[N];
long long x;
long long cal(long long n) {
  n--;
  long long tmp = n * (n + 1) * (2 * n + 1) / 6;
  n++;
  tmp -= (n * n * n - n * n) / 2;
  if (2 * (x - tmp) % (n * n + n)) return -1;
  long long m = 2 * (x - tmp) / (n * n + n);
  return m;
}
void solve() {
  int cnt = 0, tot = 0;
  for (int i = 1; i < 1500000; i++) {
    long long m = cal((long long)i);
    if (m == -1) continue;
    if (m < i) break;
    ans1[tot++] = make_pair(i, m);
    if (m == i) break;
    ans2[cnt++] = make_pair(m, i);
  }
  printf("%d\n", tot + cnt);
  for (int i = 0; i < tot; i++)
    printf("%I64d %I64d\n", ans1[i].first, ans1[i].second);
  for (int i = cnt - 1; i >= 0; i--)
    printf("%I64d %I64d\n", ans2[i].first, ans2[i].second);
}
int main() {
  scanf("%I64d", &x);
  solve();
  return 0;
}