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 O f N and M, such that there is exactlyx distinct squares in the table consisting ofn rows and m Columns. For example, in a3?x?5 table there is squares with Side one,8 squares with side and 3 squares with side three. The total number of distinct squares in a 3?x?5 table is 15?+?8?+?3?=?26.

Input

The first line of the input contains a single integer x (1?≤? x≤?10)-the number of squares inside the tables Spongebob is interested in.

Output

First print a single integer k -the number of tables with exactlyx distinct squares inside.

Then print k pairs of integers describing the tables. Print the pairs in the order of increasingn, and in case of equality-in the order of increasingM.

Examplesinput

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?x?2 table There is 2 1?x?1 squares. So, 2 distinct squares in total.

In a 2?x?3 table There is 6 1?x?1 squares and 22?x?2 squares. That's equal to 8 squares in total.

Test Instructions: Given n, ask you how many pairs of a, b make a*b have n squares within the matrix.


Parse: Manually derive the formula for the number of squares in the a*b matrix and then brute force enumeration.

 #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; } 

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