/*
# 问题分析
题设给定一组坐标以及对应的ID，要求计算哪个坐标离远点最近和最远，这是一道最值计算问题。
与原点的距离等于两个坐标值分别平方相加后的正平方根。即, distance = sqrt(x^2 + y^2)
 
# 完整步骤描述
1. 获取输入: 选手人数（坐标组数）
2. 初始化记录器:
    - 最近的距离 = 300
    - 最近距离的ID
    - 最远的距离 = -1
    - 最远距离的ID
3. 对于每一组坐标:
    - 计算坐标离远点距离
    - 如果距离小于当前最近距离:
        - 当前最近距离 = 当前计算得到的距离
        - 最近距离的ID = 当前ID
    - 如果距离大于当前最远距离:
        - 当前最远距离 = 当前计算得到的距离
        - 最远距离的ID = 当前ID
4. 输出最近距离的ID和最远距离的ID
 
# 伪代码描述
1. get input: player_amount
2. init recorder:
    - max_distance = -1
    - best_player_ID
    - min_distance = 300
    - worst_player_ID
3. for each player:
    - get input: ID, x, y
    - calculate distance = sqrt(x * x + y * y);
    - if distance < min_distance:
        - min_distance = distance;
        - best_player_ID = ID;
    - if distance > max_distance:
        - max_distance = distance;
        - worst_player_ID = ID
4. print(best_player_ID, worst_player_ID);
*/
 
 
# include<stdio.h>
# include<math.h>
 
int main(){
    int player_amount;
    scanf("%d", &player_amount);
    double min_distance = 300;
    double max_distance = -1;
    int best_player_ID, worst_player_ID;
    
    int player_ID, x, y;
    for (int i = 0; i < player_amount; i++){
        scanf("%d %d %d", &player_ID, &x, &y);
        double distance = sqrt(x * x + y * y);
        if (distance < min_distance){
            min_distance = distance;
            best_player_ID = player_ID;
        }
        if (distance > max_distance){
            max_distance = distance;
            worst_player_ID = player_ID;
        }
    }
    
    printf("%04d %04d", best_player_ID, worst_player_ID);
    return 0;
}