## I'm a slow walker, but I never walk backwards.

We have two classes, class 0 and class 1. Each student is in one of these two classes and has his own ID and grades (upto 100) from all 18 quizzes. We adjust grades based on the average grades from two classes. Students in the class with the lower average grade will receive extra points. We first compute the ratio (denoted by $r$) between the higher to the lower average grade from two classes. Then we adjust the grades from the class with the lower average by multiplying them by $r$. If the adjusted grades exceeds 100, then it becomes 100. We do this for every quiz. Note that the class to receive this $r$ bonus may be different from a quiz to another. Please refer to the following figure for calculating the average of a class and adjusting the score.

• average score = total score of the class / number of students of the class
• ratio = higher average score / lower average score
• scaled score = raw score * ratio

The final grade of a student is determined by 18 adjusted quiz scores. Each adjusted quiz score worths 6 points of the final grade, and there are 108 points in total. If the final grade exceeds 100, it becomes 100. Please refer the following figure for the formula for calculating the final grade of a student.

• final grade = 0.06 * the sum of the 18 exam scores

Let us illustrate the task by Sample Input 2 below.

Class 0 R07944051 R07922133
Exam 0 90 10
Exam 1 40 10
Exam 2 10 100
Exam 3 50 50
Exam 4 10 70
Exam 5 80 10
Exam 6 50 80
Exam 7 90 50
Exam 8 20 40
Exam 9 100 60
Exam 10 30 10
Exam 11 60 0
Exam 12 60 100
Exam 13 0 100
Exam 14 90 10
Exam 15 10 60
Exam 16 50 10
Exam 17 20 0
Class 1 R07944040 B03902075 R06922157 B07902136
Exam 0 40 80 70 80
Exam 1 20 90 80 90
Exam 2 20 100 70 90
Exam 3 20 20 100 90
Exam 4 30 20 80 90
Exam 5 50 40 10 90
Exam 6 60 20 70 90
Exam 7 10 80 30 90
Exam 8 50 70 90 90
Exam 9 0 80 60 90
Exam 10 40 50 10 90
Exam 11 30 30 40 90
Exam 12 90 50 70 90
Exam 13 0 100 0 90
Exam 14 10 0 40 90
Exam 15 80 40 30 90
Exam 16 100 80 60 90
Exam 17 80 50 40 100

In exam 0, the average score of class 0 is (90 + 10) / 2 = 50, and the average score of class 1 is (40 + 80 + 70 + 80) / 4 = 67.5, and the ratio $r$ is 67.5 / 50 = 1.35. Therefore, all scores in class 0 will be multiplied by 1.35. The score of ID R07922133 now becomes 13.5. The score of ID R07944051 becomes 121.5, then becomes 100 because it exceeds 100. The following tables are the adjusted score of class 0 and class 1.

Class 0 R07944051 R07922133
Exam 0 100.000000 13.500000
Exam 1 100.000000 28.000000
Exam 2 12.727273 100.000000
Exam 3 57.500000 57.500000
Exam 4 13.750000 96.250000
Exam 5 84.444444 10.555556
Exam 6 50.000000 80.000000
Exam 7 90.000000 50.000000
Exam 8 50.000000 100.000000
Exam 9 100.000000 60.000000
Exam 10 71.250000 23.750000
Exam 11 95.000000 0.000000
Exam 12 60.000000 100.000000
Exam 13 0.000000 100.000000
Exam 14 90.000000 10.000000
Exam 15 17.142857 100.000000
Exam 16 100.000000 27.500000
Exam 17 100.000000 0.000000
Class 1 R07944040 B03902075 R06922157 B07902136
Exam 0 40.000000 80.000000 70.000000 80.000000
Exam 1 20.000000 90.000000 80.000000 90.000000
Exam 2 20.000000 100.000000 70.000000 90.000000
Exam 3 20.000000 20.000000 100.000000 90.000000
Exam 4 30.000000 20.000000 80.000000 90.000000
Exam 5 50.000000 40.000000 10.000000 90.000000
Exam 6 65.000000 21.666667 75.833333 97.500000
Exam 7 13.333333 100.000000 40.000000 100.000000
Exam 8 50.000000 70.000000 90.000000 90.000000
Exam 9 0.000000 100.000000 83.478261 100.000000
Exam 10 40.000000 50.000000 10.000000 90.000000
Exam 11 30.000000 30.000000 40.000000 90.000000
Exam 12 96.000000 53.333333 74.666667 96.000000
Exam 13 0.000000 100.000000 0.000000 94.736842
Exam 14 14.285714 0.000000 57.142857 100.000000
Exam 15 80.000000 40.000000 30.000000 90.000000
Exam 16 100.000000 80.000000 60.000000 90.000000
Exam 17 80.000000 50.000000 40.000000 100.000000

After adjusting all the quiz scores, we calculate the final grade for each student.

The final score of ID R07944051 is $(100.000000+100.000000+12.727273+57.500000+13.750000+84.444444+50.000000 \\+90.000000+50.000000+100.000000+71.250000+95.000000+60.000000+0.000000 \\+90.000000+17.142857+100.000000+100.000000) \times 0.06 = 71.508874$

The final scores of ID B07902136 is $(80.000000+90.000000+90.000000+90.000000+90.000000+90.000000+97.500000 \\+100.000000+90.000000+100.000000+90.000000+90.000000+96.000000+94.736842 \\+100.000000+90.000000+90.000000+100.000000) \times 0.06 = 100.094211$ , then becomes 100 because it exceeds 100, and so on.

We define classes, student and score as follow. In Classes, numStudent is the number of students. In Student, finalScore is the student’s total score.

typedef struct{    int rawScore;    double scaledScore;}Score; typedef struct{    char id[10];    Score score[18];    double finalScore;}Student; typedef struct{    int numStudent;    Student student[1000];}Classes;


The prototype of the function you need to implement is as follows.

void computeGrade(Classes classes[2]);


And you have to use the following main function to test your function.

#include <stdio.h>#include "computeGrade.h" int main(){    Classes classes[2];    scanf("%d", &classes[0].numStudent);    for (int i = 0; i < classes[0].numStudent; i++) {        scanf("%s", classes[0].student[i].id);        for (int j = 0; j < 18; j++) {            scanf("%d", &classes[0].student[i].score[j].rawScore);        }    }    scanf("%d", &classes[1].numStudent);    for (int i = 0; i < classes[1].numStudent; i++) {        scanf("%s", classes[1].student[i].id);        for (int j = 0; j < 18; j++) {            scanf("%d", &classes[1].student[i].score[j].rawScore);        }    }    computeGrade(classes);    return 0;}


The computeGrade.h is as follow.

#ifndef COMPUTEGRADE_H_INCLUDED#define COMPUTEGRADE_H_INCLUDED typedef struct{    int rawScore;    double scaledScore;}Score; typedef struct{    char id[10];    Score score[18];    double finalScore;}Student; typedef struct{    int numStudent;    Student student[1000];}Classes; void computeGrade(Classes classes[2]); #endif // COMPUTEGRADE_H_INCLUDED


• 30 points : There is only one student in each of these two classes.
• 70 points : There are many students in each of these two classes.

## Input Format

The input contains only one test case. The first line contains one integer $N$, representing the number of students in class 0. Each of the following $N$ lines contains one string and 18 integers for student ID and score for each exam in class 0. The next line contains one integer $M$, representing the number of students in class 1. Each of the following $M$ lines contains one string and 18 integers for student ID and score for each exam in class 1.

• $N, M \lt 1000$
• The length of student ID is 9.
• All average scores are greater than 0.

## Note

To avoid errors that may occur in floating-point operations, you must calculate the scores in a specific order. When adjusting the score, you must first calculate the sum of the scores of the exam, and then divide by the number of students in the class to get the average score. When calculating the final grade, you must first calculate the sum of the quiz scores, and then multiplied it by 0.06. And you can only use double data type when calculating.

## Output Format

Use %f to print the final grade of each student according to the order of input data.

## Sample Input 1

1R07944040 40 90 40 10 30 40 50 90 30 40 40 70 30 50 60 20 30 401B18055151 70 10 70 60 30 10 70 20 30 80 10 80 80 90 80 20 90 90


## Sample Output 1

R07944040 72.000000B18055151 72.000000


## Sample Input 2

2R07944051 90 40 10 50 10 80 50 90 20 100 30 60 60 0 90 10 50 20R07922133 10 10 100 50 70 10 80 50 40 60 10 0 100 100 10 60 10 04R07944040 40 20 20 20 30 50 60 10 50 0 40 30 90 0 10 80 100 80B03902075 80 90 100 20 20 40 20 80 70 80 50 30 50 100 0 40 80 50R06922157 70 80 70 100 80 10 70 30 90 60 10 40 70 0 40 30 60 40B07902136 80 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 100


## Sample Output 2

R07944051 71.508874R07922133 57.423333R07944040 44.917143B03902075 62.700000R06922157 60.667267B07902136 100.000000