Description

“Point, point, life of student!” 

This is a ballad（歌谣）well known in colleges, and you must care about your score in this exam too. How many points can you get? Now, I told you the rules which are used in this course. 

There are 5 problems in this final exam. And I will give you 100 points if you can solve all 5 problems; of course, it is fairly difficulty for many of you. If you can solve 4 problems, you can also get a high score 95 or 90 (you can get the former(前者) only when your rank is in the first half of all students who solve 4 problems). Analogically（以此类推）, you can get 85、80、75、70、65、60. But you will not pass this exam if you solve nothing problem, and I will mark your score with 50. 

Note, only 1 student will get the score 95 when 3 students have solved 4 problems. 

I wish you all can pass the exam! 

Come on! 

 

Input

Input contains multiple test cases. Each test case contains an integer N (1<=N<=100, the number of students) in a line first, and then N lines follow. Each line contains P (0<=P<=5 number of problems that have been solved) and T（consumed time）. You can assume that all data are different when 0<p. 

A test case starting with a negative integer terminates the input and this test case should not to be processed. 

 

Output

Output the scores of N students in N lines for each case, and there is a blank line after each case. 

 

Sample Input

4 5 06:30:17 4 07:31:27 4 08:12:12 4 05:23:13 1 5 06:30:17 -1

 

Sample Output

100 90 90 95 100

 

1 #include
       
         2 #include
        
          3 #include
         
           4 #include
          
            5 using namespace std; 6  7 int main() 8 { 9     int n;10     struct 11     {12             char t[10];//time13             int p;//problem14             int s;//score15     }a[120];16     while((scanf("%d",&n))&&n!=-1)17     {18         for(int i=0;i
           
            0)41                     {42                         char p[10];43                         strcpy(p,time[i]);44                         strcpy(time[i],time[j]);45                         strcpy(time[j],p);46                     }47                 }48             }49             for(int i=0;i