题干：

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

解题报告：

  有坑就是不能只判断行和列是否相同，还需要判断行和列是否重合了，比如全输入0 0。所以需要“//”的那两行。

  所以以后要注意这种坑啊！重合问题。还有考虑全0的这种特判。

AC代码：

#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