#include <bits/stdc++.h>
using namespace std;
const long double eps = 1e-9;
const long double PI = atan2((long double)0, (long double)-1);
const long long MOD = 1000000007;
const int INF = 1 << 30;
const int ARR_INF = 1 << 6;
const int MAXN = 1642250;
long long x;
long long sumsqr[MAXN];
vector<pair<long long, long long> > vec;
void Init() {
  sumsqr[0] = 0;
  for (long long i = 1; i < MAXN; i++) sumsqr[i] = sumsqr[i - 1] + i * i;
  return;
}
long long Getsum(long long n) { return n * (n + 1) / 2; }
int main() {
  Init();
  while (scanf("%I64d", &x) != EOF) {
    vec.clear();
    for (long long i = 1; i < MAXN; i++) {
      if (sumsqr[i] > x) break;
      long long sum = Getsum(i);
      long long tms = (x - sumsqr[i]) / sum;
      long long xx = sumsqr[i] + sum * tms;
      if (xx == x) vec.push_back(make_pair(i, i + tms));
    }
    sort(vec.begin(), vec.end());
    int siz = vec.size();
    int plu = 0;
    if (vec[siz - 1].first == vec[siz - 1].second) plu = -1;
    printf("%d\n", 2 * siz + plu);
    for (int i = 0; i < siz; i++)
      printf("%I64d %I64d\n", vec[i].first, vec[i].second);
    for (int i = siz - 1 + plu; i >= 0; i--)
      printf("%I64d %I64d\n", vec[i].second, vec[i].first);
  }
  return 0;
}

### Prompt

Your challenge is to write a Cpp solution to the following 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.

<image>

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>

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const long double eps = 1e-9;
const long double PI = atan2((long double)0, (long double)-1);
const long long MOD = 1000000007;
const int INF = 1 << 30;
const int ARR_INF = 1 << 6;
const int MAXN = 1642250;
long long x;
long long sumsqr[MAXN];
vector<pair<long long, long long> > vec;
void Init() {
  sumsqr[0] = 0;
  for (long long i = 1; i < MAXN; i++) sumsqr[i] = sumsqr[i - 1] + i * i;
  return;
}
long long Getsum(long long n) { return n * (n + 1) / 2; }
int main() {
  Init();
  while (scanf("%I64d", &x) != EOF) {
    vec.clear();
    for (long long i = 1; i < MAXN; i++) {
      if (sumsqr[i] > x) break;
      long long sum = Getsum(i);
      long long tms = (x - sumsqr[i]) / sum;
      long long xx = sumsqr[i] + sum * tms;
      if (xx == x) vec.push_back(make_pair(i, i + tms));
    }
    sort(vec.begin(), vec.end());
    int siz = vec.size();
    int plu = 0;
    if (vec[siz - 1].first == vec[siz - 1].second) plu = -1;
    printf("%d\n", 2 * siz + plu);
    for (int i = 0; i < siz; i++)
      printf("%I64d %I64d\n", vec[i].first, vec[i].second);
    for (int i = siz - 1 + plu; i >= 0; i--)
      printf("%I64d %I64d\n", vec[i].second, vec[i].first);
  }
  return 0;
}

```