Codeforces Round #332 (Div. 2) D. Spongebob and Squares

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.

In a 2 × 3 table there are 6 1 × 1 squares and 2 2 × 2 squares. That is equal to 8 squares in total.

题意：给定n，问你有存在多少对a,b使得a*b的矩阵内有n个正方形。


分析：手动推导一下a*b的矩阵中正方形个数的公式然后暴力枚举即可。

#include <cstdio>
#include <iostream>
using namespace std;
struct thing
{
	long long x,y;
} ans[1000000];
long long a,n,tot,same;
int main()
{
	cin.sync_with_stdio(false);
	cin>>n;
	long long i = 1;
	while(true)
	{
		long long b = (i-1)*i*(2*i-1)/6 - i*i*(i-1)/2;
		if(b >= n) break;
		long long k = i*(i+1)/2;
		if((n - b) / k < i) break;
		if((n - b) % k == 0) 
		{
			ans[++tot].x = i;
			ans[tot].y = (n - b) / k;
			if(ans[tot].x == ans[tot].y) same++;
		}
		i++;
	}
	cout<<2*tot-same<<endl;
	for(int i = 1;i <= tot;i++) cout<<ans[i].x<<" "<<ans[i].y<<endl;
	for(int i = tot;i;i--) 
	 if(ans[i].x != ans[i].y)
	  cout<<ans[i].y<<" "<<ans[i].x<<endl; 
 }