題干：

Gerald is very particular to eight point sets. He thinks that any decent eight point set must consist of all pairwise intersections of three distinct integer vertical straight lines and three distinct integer horizontal straight lines, except for the average of these nine points. In other words, there must be three integers?x1,?x2,?x3and three more integers?y1,?y2,?y3, such that?x1?<?x2?<?x3,?y1?<?y2?<?y3?and the eight point set consists of all points?(xi,?yj)?(1?≤?i,?j?≤?3), except for point?(x2,?y2).

You have a set of eight points. Find out if Gerald can use this set?

Input

The input consists of eight lines, the?i-th line contains two space-separated integers?xi?and?yi?(0?≤?xi,?yi?≤?106). You do not have any other conditions for these points.

Output

In a single line print word "respectable", if the given set of points corresponds to Gerald's decency rules, and "ugly" otherwise.

Examples

Input

0 0 0 1 0 2 1 0 1 2 2 0 2 1 2 2

Output

respectable

Input

0 0 1 0 2 0 3 0 4 0 5 0 6 0 7 0

Output

ugly

Input

1 1 1 2 1 3 2 1 2 2 2 3 3 1 3 2

Output

ugly

解題報告：

? 有坑就是不能只判斷行和列是否相同，還需要判斷行和列是否重合了，比如全輸入0 0。所以需要“//”的那兩行。

? 所以以后要注意這種坑啊！重合問題。還有考慮全0的這種特判。

AC代碼：

#include<bits/stdc++.h> #define ll long long using namespace std;struct Node {int x,y;int id; } n[9]; int ck[3] = {1,4,6}; int ck2[3] = {3,5,8}; bool cmp(Node a,Node b) {if(a.x != b.x) return a.x < b.x;else return a.y < b.y; } int main() {for(int i = 1; i<=8; i++) {scanf("%d%d",&n[i].x,&n[i].y);}sort(n+1,n+8+1,cmp);int flag = 1 ;for(int i = 1; i<3; i++) {if(n[i].x != n[i+1].x) flag=0;if(n[i].y == n[i+1].y) flag=0;//}if(n[4].x != n[5].x) flag=0;for(int i = 6; i<8; i++) {if(n[i].x != n[i+1].x) flag=0;}for(int k = 0; k<2; k++) {if(n[ck[k]].y != n[ck[k+1]].y) flag=0;if(n[ck[k]].x == n[ck[k+1]].x) flag=0;//}if(n[2].y != n[7].y) flag=0;for(int k = 0; k<2; k++) {if(n[ck2[k]].y != n[ck2[k+1]].y) flag=0;}if(flag == 1) puts("respectable");else puts("ugly");return 0 ; } //0 0 //0 1 //0 2 //1 0 //1 2 //1 0 //1 1 //1 2

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