#include <bits/stdc++.h>
using namespace std;
struct Point {
    int x, y;
};
int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int N;
    cin >> N;
    int total = 2 * N;
    vector<Point> pts(total);
    for (int i = 0; i < total; i++){
        cin >> pts[i].x >> pts[i].y;
    }
    // ① 先尝试竖直直线方案：按 x 坐标排序（当 x 相同时按 y 排序）
    {
        vector<Point> temp = pts;
        sort(temp.begin(), temp.end(), [](const Point &a, const Point &b){
            return a.x == b.x ? a.y < b.y : a.x < b.x;
        });
        int x_low = temp[N-1].x, x_high = temp[N].x;
        if(x_low < x_high){
            if(x_low + 1 < x_high){
                int t = x_low + 1;
                cout << 1 << " " << 0 << " " << -t << "\n";
                return 0;
            } else {
                int t = 2 * x_low + 1;
                cout << 2 << " " << 0 << " " << -t << "\n";
                return 0;
            }
        }
    }
    // ② 当竖直方案不可行，则必有部分点共享相同 x 值，此时尝试水平直线方案：按 y 坐标排序（当 y 相同时按 x 排序）
    {
        vector<Point> temp = pts;
        sort(temp.begin(), temp.end(), [](const Point &a, const Point &b){
            return a.y == b.y ? a.x < b.x : a.y < b.y;
        });
        int y_low = temp[N-1].y, y_high = temp[N].y;
        if(y_low < y_high){
            if(y_low + 1 < y_high){
                int t = y_low + 1;
                cout << 0 << " " << 1 << " " << -t << "\n";
                return 0;
            } else {
                int t = 2 * y_low + 1;
                cout << 0 << " " << 2 << " " << -t << "\n";
                return 0;
            }
        }
    }
    // ③ 按理来说至少有一种方案必定可行，故此处理论上不会到达
    cout << "1 0 0\n";
    return 0;
}