A - Nias and Tug-of-War

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

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Description

Nias is fond of tug-of-war. One day, he organized a tug-of-war game and invited a group of friends to take part in.

Nias will divide them into two groups. The strategy is simple, sorting them into a row according to their height from short to tall, then let them say one and two alternately (i.e. one, two, one, two...). The people who say one are the members of the red team while others are the members of the blue team.

We know that the team which has a larger sum of weight has advantages in the tug-of-war. Now give you the guys' heights and weights, please tell us which team has advantages.

Input

The first line of input contains an integer T, indicating the number of test cases.

The first line of each test case contains an integer N (N is even and 6 ≤ N ≤ 100). 

Each of the next N lines contains two real numbers X and Y, representing the height and weight of a friend respectively.

Output

One line for each test case. If the red team is more likely to win, output "red", if the blue team is more likely to win, output "blue". If both teams have the same weight, output "fair".

Sample Input

1 6 170 55 165.3 52.5 180.2 60.3 173.3 62.3 175 57 162.2 50

Sample Output

blue

【分析】

题意：一群人按身高排序，然后奇数去blue偶数去red，两边的人体重加起来重的一方赢，问哪边能赢....

排个序....比较

【代码】


#include <stdio.h>
#include <algorithm>
using namespace std;
struct xx{
  double x,y;
}a[100000];

int cmp(const xx &q,const xx &w)
{
  return q.x<w.x;
}


int main()
{
  int pp;scanf("%d",&pp);
  while (pp--)
  {
    int n;scanf("%d",&n);
    for (int i=0;i<n;i++) scanf("%lf%lf",&a[i].x,&a[i].y);
    sort(a,a+n,cmp);
    double ans1=0;
    double ans2=0;
    for (int i=0;i<n;i++)
      if (i%2)
        ans1+=a[i].y;
      else
        ans2+=a[i].y; 
    if (ans1==ans2) printf("fair\n");
    if (ans1<ans2) printf("red\n");
    if (ans1>ans2) printf("blue\n");
  }
}