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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <iostream>
#include <cstdio>
#include <string>
#include <vector>
#include <algorithm>
#include <functional>
#include <iomanip>
#include <stdlib.h>
#include <string.h>
#include <cstring>
#include <cmath>
#include <map>
#include <queue>
#include <deque>
#include <stack>
#include <set>
#include <stdio.h>
#include <tuple>
#include <bitset>
#include <cfloat>
#include <fstream>
#include <limits.h>
#include <list>
#include <math.h>
#include <random>
#include <unordered_map>
#include <unordered_set>
#include <bitset>
#include <numeric>
#include <utility>
#include <sstream>
#include <cassert>
template<class T> inline bool chmin(T& a, T b) {
	if (a > b) {
		a = b;
		return true;
	}
	return false;
}
template<class T> inline bool chmax(T& a, T b) {
	if (a < b) {
		a = b;
		return true;
	}
	return false;
}
using namespace std;
#define int long long
#define ll long long
#define rep(i, n) for (ll i = 0; i < (n); i++)
#define P pair<ll, ll>
#define sz(x) (ll)x.size()
#define ALL(x) (x).begin(),(x).end()
#define ALLR(x) (x).rbegin(),(x).rend()
#define VE vector<ll>
#define COUT(x) cout<<(x)<<endl
#define MA map<ll,ll>
#define SE set<ll>
#define PQ priority_queue<ll>
#define PQR priority_queue<ll,VE,greater<ll>>
#define YES(n) cout << ((n) ? "YES" : "NO"  ) << endl
#define Yes(n) cout << ((n) ? "Yes" : "No"  ) << endl
#define EPS (1e-14)
#define pb push_back
long long MOD = 1000000007;
//const long long MOD = 998244353;
const long long INF = 1LL << 60;
const double PI = acos(-1.0);
struct mint {
	ll x; // typedef long long ll;
	mint(ll x = 0) :x((x%MOD + MOD) % MOD) {}
	mint operator-() const { return mint(-x); }
	mint& operator+=(const mint a) {
		if ((x += a.x) >= MOD) x -= MOD;
		return *this;
	}
	mint& operator-=(const mint a) {
		if ((x += MOD - a.x) >= MOD) x -= MOD;
		return *this;
	}
	mint& operator*=(const mint a) {
		(x *= a.x) %= MOD;
		return *this;
	}
	mint operator+(const mint a) const {
		mint res(*this);
		return res += a;
	}
	mint operator-(const mint a) const {
		mint res(*this);
		return res -= a;
	}
	mint operator*(const mint a) const {
		mint res(*this);
		return res *= a;
	}
	mint pow(ll t) const {
		if (!t) return 1;
		mint a = pow(t >> 1);
		a *= a;
		if (t & 1) a *= *this;
		return a;
	}
	// for prime MOD
	mint inv() const {
		return pow(MOD - 2);
	}
	mint& operator/=(const mint a) {
		return (*this) *= a.inv();
	}
	mint operator/(const mint a) const {
		mint res(*this);
		return res /= a;
	}
};
istream& operator>>(istream& is, const mint& a) { return is >> a.x; }
ostream& operator<<(ostream& os, const mint& a) { return os << a.x; }
struct combination {
	vector<mint> fact, ifact;
	combination(int n) :fact(n + 1), ifact(n + 1) {
		assert(n < MOD);
		fact[0] = 1;
		for (int i = 1; i <= n; ++i) fact[i] = fact[i - 1] * i;
		ifact[n] = fact[n].inv();
		for (int i = n; i >= 1; --i) ifact[i - 1] = ifact[i] * i;
	}
	mint operator()(int n, int k) {
		if (k < 0 || k > n) return 0;
		return fact[n] * ifact[k] * ifact[n - k];
	}
};
struct Sieve {
	int n;
	vector<int> f, primes;

	Sieve(int n = 1) :n(n), f(n + 1) {
		f[0] = f[1] = -1;
		for (ll i = 2; i <= n; ++i) {
			if (f[i]) continue;
			primes.push_back(i);
			f[i] = i;
			for (ll j = i * i; j <= n; j += i) {
				if (!f[j]) f[j] = i;
			}
		}
	}

	bool isPrime(int x) { return f[x] == x; }

	vector<int> factorList(int x) {
		vector<int> res;
		while (x != 1) {
			res.push_back(f[x]);
			x /= f[x];
		}
		return res;
	}

	vector<P> factor(int x) {
		vector<int> fl = factorList(x);
		if (fl.size() == 0) return {};
		vector<P> res(1, P(fl[0], 0));
		for (int p : fl) {
			if (res.back().first == p) {
				res.back().second++;
			}
			else {
				res.emplace_back(p, 1);
			}
		}
		return res;
	}
};
template<class t> t gcd(t a, t b) { return b != 0 ? gcd(b, a % b) : a; }
ll lcm(ll a, ll b) {
	ll g = gcd(a, b);
	return a / g * b;
}
bool prime(ll n) {
	for (ll i = 2; i <= sqrt(n); i++) {
		if (n%i == 0)return false;
	}
	return n != 1;
}
map<ll, ll> prime_factor(ll n) {
	map<ll, ll> ret;
	for (ll i = 2; i * i <= n; i++) {
		while (n % i == 0) {
			ret[i]++;
			n /= i;
		}
	}
	if (n != 1) ret[n] = 1;
	return ret;
}
ll modinv(ll a, ll m) {
	ll b = m, u = 1, v = 0;
	while (b) {
		ll t = a / b;
		a -= t * b; swap(a, b);
		u -= t * v; swap(u, v);
	}
	u %= m;
	if (u < 0) u += m;
	return u;
}
vector<pair<char, int>> RunLength(string s) {
	if (s.size() == 0)return {};
	vector<pair<char, int>>res(1, pair<char, int>(s[0], 0));
	for (char p : s) {
		if (res.back().first == p) {
			res.back().second++;
		}
		else {
			res.emplace_back(p, 1);
		}
	}
	return res;
}
// Digit Count
int GetDigit(int num) {
	return log10(num) + 1;
}
// bit calculation[how many "1"] (= __builtin_popcount())
int bit_count(int n) {
	int cnt = 0;
	while (n > 0) {
		if (n % 2 == 1)cnt++;
		n /= 2;
	}
	return cnt;
}
vector<long long> enum_divisors(long long N) {
	vector<long long> res;
	for (long long i = 1; i * i <= N; ++i) {
		if (N % i == 0) {
			res.push_back(i);
			if (N / i != i) res.push_back(N / i);
		}
	}
	sort(res.begin(), res.end());
	return res;
}
const ll dx[4] = { 1,0,-1,0 };
const ll dy[4] = { 0,1,0,-1 };
struct edge { ll to, cost; };
typedef long double ld;
using Graph = vector<VE>;
class UnionFind {
public:
	vector <ll> par; // 各元の親を表す配列
	vector <ll> siz; // 素集合のサイズを表す配列(1 で初期化)

	// Constructor
	UnionFind(ll sz_) : par(sz_), siz(sz_, 1) {
		for (ll i = 0; i < sz_; ++i) par[i] = i; // 初期では親は自分自身
	}
	void init(ll sz_) {
		par.resize(sz_);
		siz.resize(sz_, 1);
		for (ll i = 0; i < sz_; ++i) par[i] = i; // 初期では親は自分自身
	}

	// Member Function
	// Find
	ll root(ll x) { // 根の検索
		while (par[x] != x) {
			x = par[x] = par[par[x]]; // x の親の親を x の親とする
		}
		return x;
	}

	// Union(Unite, Merge)
	bool merge(ll x, ll y) {
		x = root(x);
		y = root(y);
		if (x == y) return false;
		// merge technique（データ構造をマージするテク．小を大にくっつける）
		if (siz[x] < siz[y]) swap(x, y);
		siz[x] += siz[y];
		par[y] = x;
		return true;
	}

	bool issame(ll x, ll y) { // 連結判定
		return root(x) == root(y);
	}

	ll size(ll x) { // 素集合のサイズ
		return siz[root(x)];
	}
};


signed main() {
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	//cout << fixed << setprecision(15);
	//combination com(200010);
	int n; cin >> n;
	vector<P> ans;
	if (n % 2) {
		for (int i = 1; i < n - 1; i++) {
			for (int j = n - 1; j > i; j--) {
				if (i + j == n)continue;
				ans.push_back(make_pair(i, j));
			}
		}
		rep(i, n) {
			if (i == 0)continue;
			ans.push_back(make_pair(i, n));
		}cout << ans.size() << endl;
		rep(i, ans.size())cout << ans[i].first << " " << ans[i].second << endl;
	}
	else {
		for (int i = 1; i <= n - 1; i++) {
			for (int j = i + 1; j <= n; j++) {
				if (i + j == n + 1)continue;
				ans.push_back(make_pair(i, j));
			}
		}
		cout << ans.size() << endl;
		rep(i, ans.size())cout << ans[i].first << " " << ans[i].second << endl;
	}
	return 0;
}