// [想定解答 3]
//    - 4 手読みをして最もスコアの高いものを出力する．
//    - 具体的には，組 (i, j, k, l) を全探索して，カード (i, j) -> (j, k) -> (k, l) -> (l, 1) の順に操作を行う．
//    - 25% 程度の得点が得られる．

#include <iostream>
#include <cmath>
#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];

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

	// Step 2. 誤差の計算
	long long MinV = (1LL << 60);
	vector<pair<int, int>> Answer;
	for (int i = 2; i <= N; i++) {
		for (int j = 2; j <= N; j++) {
			for (int k = 2; k <= N; k++) {
				for (int l = 2; l <= N; l++) {
					if (i == j || j == k || k == l) continue;
					for (int m = 1; m <= N; m++) C[m] = A[m];
					for (int m = 1; m <= N; m++) D[m] = B[m];
					long long avgC1 = (C[i] + C[j]) / 2LL;
					long long avgD1 = (D[i] + D[j]) / 2LL;
					C[i] = avgC1; D[i] = avgD1;
					C[j] = avgC1; D[j] = avgD1;
					long long avgC2 = (C[j] + C[k]) / 2LL;
					long long avgD2 = (D[j] + D[k]) / 2LL;
					C[j] = avgC2; D[j] = avgD2;
					C[k] = avgC2; D[k] = avgD2;
					long long avgC3 = (C[k] + C[l]) / 2LL;
					long long avgD3 = (D[k] + D[l]) / 2LL;
					C[k] = avgC3; D[k] = avgD3;
					C[l] = avgC3; D[l] = avgD3;
					long long avgC4 = (C[l] + C[1]) / 2LL;
					long long avgD4 = (D[l] + D[1]) / 2LL;
					C[l] = avgC4; D[l] = avgD4;
					C[1] = avgC4; D[1] = avgD4;
					long long err = max(abs(C[1] - Target), abs(D[1] - Target));
					if (MinV > err) {
						MinV = err;
						Answer.clear();
						Answer.push_back(make_pair(i, j));
						Answer.push_back(make_pair(j, k));
						Answer.push_back(make_pair(k, l));
						Answer.push_back(make_pair(l, 1));
					}
				}
			}
		}
	}

	// Step 3. 出力
	cout << Answer.size() << endl;
	for (int i = 0; i < Answer.size(); i++) cout << Answer[i].first << " " << Answer[i].second << endl;
	return 0;
}