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.

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Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long int a[1000000][2];
struct node {
  long long int x, y;
} b[1000000];
bool cmp(node a, node b) { return a.x < b.x; }
int main() {
  long long int i, j, k, m, n, x, y, z, mark, temp;
  while (~scanf("%lld", &x)) {
    memset(a, 0, sizeof(a));
    memset(b, 0, sizeof(b));
    j = 0;
    mark = 0;
    k = (long long int)sqrt(x);
    if (x == 1) mark = 1;
    for (i = 1; i <= 20000000 && i <= k; i++) {
      y = x - i * (i - 1) * (2 * i - 1) / 6 + i * i * (i - 1) / 2;
      z = i * i - i * (i - 1) / 2;
      if (z == 0) {
        a[j][0] = i;
        a[j++][1] = x;
      } else if (y % z == 0 && i <= y / z) {
        a[j][0] = i;
        a[j++][1] = y / z;
        if (a[j - 1][0] == a[j - 1][1]) mark = 1;
      }
    }
    k = 2 * j;
    if (mark) k--;
    printf("%lld\n", k);
    k = 0;
    for (i = 0; i < j; i++) {
      b[k].x = a[i][0];
      b[k].y = a[i][1];
      k++;
    }
    if (!mark) {
      b[k].x = a[j - 1][1];
      b[k].y = a[j - 1][0];
      k++;
    }
    for (i = j - 2; i >= 0; i--) {
      b[k].x = a[i][1];
      b[k].y = a[i][0];
      k++;
    }
    sort(b, b + k, cmp);
    for (i = 0; i < k; i++) printf("%lld %lld\n", b[i].x, b[i].y);
  }
  return 0;
}