Problem:
Score : 700 points

Problem Statement

You are given an integer N.
Build an undirected graph with N vertices with indices 1 to N that satisfies the following two conditions:

The graph is simple and connected.

There exists an integer S such that, for every vertex, the sum of the indices of the vertices adjacent to that vertex is S.

It can be proved that at least one such graph exists under the constraints of this problem.

Constraints

All values in input are integers.

3 \leq N \leq 100

Input

Input is given from Standard Input in the following format:

N

Output

In the first line, print the number of edges, M, in the graph you made. In the i-th of the following M lines, print two integers a_i and b_i, representing the endpoints of the i-th edge.

The output will be judged correct if the graph satisfies the conditions.

Sample Input 1

3

Sample Output 1

2
1 3
2 3

For every vertex, the sum of the indices of the vertices adjacent to that vertex is 3.

Solution in Ruby (2.3.3) (judge verdict: Accepted)

class UnionFind
  def initialize(n); @data = Array.new(n,-1); end
  def find(a); @data[a] < 0 ? a : @data[a] = find(@data[a]); end
  def same?(a,b); find(a) == find(b); end
  def size(a); -@data[find(a)]; end
  def unite(a,b)
    a = find(a); b = find(b)
    return if a == b
    a,b = b,a if @data[a] > @data[b]
    @data[a] += @data[b]; @data[b] = a
  end
end

# UnionFind.new(N) := 要素数Nで初期化
# Ai := i番目の要素
# find(i) := Aiが属するグループ
# same?(i,j) := Ai,Ajが同じグループに属する
# size(i) := Aiが属するグループの要素の数
# unite(i,j) := Ai,Ajを同じグループに統合する

n = gets.to_i
uf = UnionFind.new(n+1)

1.upto(n/2) do |i|
  j = n - i
  j += 1 if n.even?
  uf.unite(i,j)
end

g = []
1.upto(n-1) do |i|
  (i+1).upto(n) do |j|
    next if uf.same?(i,j)
    g << [i,j].join(" ")
  end
end

puts g.size
g.each do |v|
  puts v
end