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 PyPy3 (2.4.0) (judge verdict: Wrong Answer)

#AGC032-B Balanced Neighbors
"""
まずnが偶数の場合を考える。
1...nの数列を考えた時に、
1+n,2+n-1,3+n-2…を１頂点として考えると、全てその値が等しいグラフになる。
次に奇数の場合を考える。
n-1までのグラフを偶数の場合と同じように構築した後、
nはn-1までの2頂点の合計と等しい値になる。
よって、最後に全頂点をnとつなぎ合わせると条件を満たすグラフに成る。
"""
import sys
from collections import defaultdict
dic1 = defaultdict(int)
readline = sys.stdin.buffer.readline
def even(n): return 1 if n%2==0 else 0
n = int(readline())
mid = 1+(n//2)*2
for i in range(1,(n//2)*2+1):
    dic1[i] = mid-i
for i in range(1,(n//2)*2):
    res = dic1[i]
    for j in range(i+1,(n//2)*2+1):
        if j == res:
            continue
        print(i,j)

if not even(n):
    for i in range(1,n):
        print(n,i)