// [想定解答 4-3]
//    - 想定解法 4-2 に加えて，スコアが小さいときに焼きなまし法の近傍を小さくする．
//    - 80% 以上の得点が得られる．

#include <iostream>
#include <cmath>
#include <ctime>
#include <vector>
#include <algorithm>
using namespace std;

const long long Target = 500000000000000000LL;
long long N;
long long A[59], C[59];
long long B[59], D[59];
long long Order[59];
long long BestOrder[59];

long long GetScore() {
	for (int i = 1; i <= N; i++) C[i] = A[i];
	for (int i = 1; i <= N; i++) D[i] = B[i];
	for (int i = 1; i <= N - 1; i++) {
		long long avgC = (C[Order[i]] + C[Order[i + 1]]) / 2LL;
		long long avgD = (D[Order[i]] + D[Order[i + 1]]) / 2LL;
		C[Order[i]] = avgC; C[Order[i + 1]] = avgC;
		D[Order[i]] = avgD; D[Order[i + 1]] = avgD;
	}
	return max(abs(Target - C[1]), abs(Target - D[1]));
}

long double Randouble() {
	double s = 0.0, t = 1.0;
	for (int i = 0; i < 3; i++) {
		t /= 1024.0;
		s += 1.0 * (rand() % 1024) * t;
	}
	return s;
}

int main() {
	// Step 1. 入力
	cin >> N;
	for (int i = 1; i <= N; i++) cin >> A[i] >> B[i];

	// Step 2. 山登り法
	for (int i = 1; i <= N; i++) Order[i] = i;
	long long CurrentScore = GetScore();
	long long BestScore = CurrentScore;
	int StartTime = clock();
	while (clock() - StartTime <= 85 * CLOCKS_PER_SEC / 100) {
		int idx1 = rand() % N + 1;
		int idx2 = (BestScore <= 100000LL ? min((int)N - 1, idx1 + rand() % 8 + 1) : rand() % N + 1);
		swap(Order[idx1], Order[idx2]);
		long long CandScore = GetScore();
		double Diff = log10(CurrentScore) - log10(CandScore);
		if (Randouble() < exp(Diff / 0.2)) {
			if (BestScore > CandScore) {
				BestScore = CandScore;
				for (int i = 1; i <= N; i++) BestOrder[i] = Order[i];
			}
			CurrentScore = CandScore;
		}
		else {
			swap(Order[idx1], Order[idx2]);
		}
	}

	// Step 3. 答えを得る
	vector<pair<int, int>> Answer;
	for (int i = 1; i <= N - 1; i++) Answer.push_back(make_pair(BestOrder[i], BestOrder[i + 1]));

	// Step 4. 出力
	cout << Answer.size() << endl;
	for (int i = 0; i < Answer.size(); i++) cout << Answer[i].first << " " << Answer[i].second << endl;
	return 0;
}