Codeforces Round #332 (Div. 2)-D Spongebob and Squares（枚举+递推）

D. Spongebob and Squares

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 15squares 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 is15 + 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.

题意：给你一个x，问你能组成多少种n*m的方格，使得铺上任意大小的正方形能铺满的方案数正好为x。

题解：我们枚举长和宽的话，复杂度一定会炸，我们可以考虑先假设n=m，呢此时是一个正方形方格，枚举出前几种方案数会发现规律，其实就是前n项的平方和，也就是tmp=1^2+2^2+...n^2=n*(n+1)*(2*n+1)/6.但此时并不一定能满足方案数正好为x。呢么就开始一层层加，每加一层你会发现方案数会多y=n+n-1+n-2+...+1=n*(n+1)/2，因此我们只需再判断（x-tmp）%y能否除尽即可，总复杂度为1e6

#include<set>      
#include<map>         
#include<stack>                
#include<queue>                
#include<vector>        
#include<string>     
#include<time.h>    
#include<math.h>                
#include<stdio.h>                
#include<iostream>                
#include<string.h>                
#include<stdlib.h>        
#include<algorithm>       
#include<functional>        
using namespace std;                
#define ll long long          
#define inf 1000000000           
#define mod 1000000007                
#define maxn  205000    
#define lowbit(x) (x&-x)                
#define eps 1e-9  
vector<pair<ll,ll> >q;
int main(void)
{
	ll n,i,tmp;
	scanf("%lld",&n);
	for(i=1;;i++)
	{
		tmp=i*(i+1)*(2*i+1)/6;
		if(tmp>n) break;
		ll now=n-tmp;
		ll x=i*(i+1)/2;
		if(now%x==0)
		{
			ll j=now/x;
			q.push_back(make_pair(i,i+j));
			if(j>0)
				q.push_back(make_pair(i+j,i));
		}
	}
	printf("%d\n",q.size());
	sort(q.begin(),q.end());
	for(i=0;i<q.size();i++)
		printf("%lld %lld\n",q[i].first,q[i].second);
	return 0;
}