Problem 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

 

Sample Output

#include<cstdio>

#include<cstring>

#include<algorithm>

#include<cmath>

using namespace std;

const int INF=0x3f3f3f3f;

#define mem(x,y) memset(x,y,sizeof(x))

#define SI(x) scanf("%d",&x)

#define PI(x) printf("%d",&x)

#define P_ printf(" ")

int ans[1010];

int nu[6];

struct Node{

	int cnt;

	int num;

	int hh,mm,dd;

	int sec;

	Node init(int x){

		cnt=x;

		scanf("%d",&num);

		nu[num]++;

		scanf("%d:%d:%d",&hh,&mm,&dd);

		sec=hh*60*60+mm*60+dd;

		ans[cnt]=50+10*num;

	}

	friend bool operator < (Node a,Node b){

		if(a.num!=b.num)return a.num<b.num;

		else return a.sec<b.sec;

	}

}dt[1010];

int main(){

	int N;

	while(SI(N),N!=-1){

		mem(nu,0);

		for(int i=0;i<N;i++){

			dt[i].init(i);

		}

		sort(dt,dt+N);

		for(int i=1;i<=4;i++){

			for(int j=0;j<N;j++){

				if(dt[j].num==i){

					if(nu[i]==1)ans[dt[j].cnt]+=5;

					else for(int k=j;k<j+nu[i]/2;k++){

						ans[dt[k].cnt]+=5;

					}

					break;

				}

			}

		}

		for(int i=0;i<N;i++){

			printf("%d\n",ans[i]);

		}

		puts("");

	}

	return 0;

}