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

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

61 262 93 55 39 226 1

input

2

output

21 22 1

input

8

output

41 82 33 28 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：给出一个整数N问有多少个n*m的格子内的正方形格子的数目等于N并输出这n*m个格子一个n*m的格子中设n为较小的一个则有这n*m的格子中正方形的数目为

n*m+(n-1)*(m-1)*...1*(m-n+1);即为n*n*m+n*(n-1)*n/2-n*(n-1)*m+1^2+2^2+....(n-1)^2;因此枚举n求出m即可。

/* ***********************************************Author       : rycCreated Time : 2016-08-23 TuesdayFile Name    : E:\acm\codeforces\332D.cppLanguage     : c++Copyright 2016 ryc All Rights Reserved************************************************ */#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<algorithm>#include<cmath>#include<list>#include<vector>#include<list>#include<queue>#include<map>#include<set>using namespace std;typedef long long LL;typedef pair<LL,LL>pll;const int maxn=1000010;set<pll>ans;set<pll>::iterator it;int main(){    LL n,sum=0;cin>>n;    ans.insert(make_pair(1,n));    ans.insert(make_pair(n,1));    for(LL i=2ll;i<=n;++i){        sum=sum+(i-1)*(i-1);        if(sum>n)break;        LL num=i*(i*(i-1)/2ll);        if((n-sum+num)%((i*i+i)/2ll))continue;        if(n-sum+num<0)break;        if(i>((n-sum+num)/((i*i+i)/2ll)))break;        ans.insert(make_pair(i,(n-sum+num)/((i*i+i)/2ll)));        ans.insert(make_pair((n-sum+num)/((i*i+i)/2ll),i));    }    printf("%d\n",ans.size());    for(it=ans.begin();it!=ans.end();++it){        printf("%lld %lld\n",(*it).first,(*it).second);    }    return 0;}