Oliver and Nova are true lovers. Inspite of knowing that Nova will die Oliver married her at the lake where they met. But they had a conflict about even and odd numbers. Nova likes the odd numbers and Oliver prefers even. One day they went to a fair where Oliver bought some square shaped marshmallows and Nova bought some round shaped. Then they decided to play a game. They will pick a natural number N . Nova will sum up the odd numbers from 1 to N and and she will notedown LCM of R(R is defined in the picture) and the sum she calculated before. And Oliver will sum up the even numbers from 1 to N and and he will notedown LCM of S(S is defined in the picture) and the sum he calculated before. You must use the ceil value of R and S.

Now  whose LCM is strictly greater than the other will win.If both of their LCM is equal Nova will win because Oliver is afraid of Nova.
$N.B.$ define the value of pi with $acos(-1)$.
$N.B.$ Sum of all odd number and sum of all even number will not exceed 10^18. 

-----Input:-----
The first line contains an integer $T$  — the number of test cases in the input. Next, T test cases are given, one per line.
Each test case is a positive integer $N$  . 

-----Output:-----
Print T answers to the test cases.
In each test cases,
If Oliver wins the game, print "Nova's gonna kill me" (without quotes) .
If Nova wins the game, print "YESS(sunglass emo)" (without quotes) .

-----Constraints-----
- $1 \leq T \leq 2000$
- $1 \leq N \leq 1845271$

-----Sample Input:-----
1
111

-----Sample Output:-----
YESS(sunglass emo)

Source: codechef | Difficulty: UNKNOWN_DIFFICULTY


Solutions:
import math

def lcm(a, b):
	return a * b // gcd(a, b)

def gcd(a, b):
	if b == 0:
		return a
	return gcd(b, a % b)
for _ in range(int(input())):
	n = int(input())
	na = math.ceil(2 * n / math.acos(-1))
	nb = ((n + 1) // 2) ** 2
	nlcm = lcm(na, nb)
	oa = math.ceil(n / 2)
	ob = n // 2 * (n // 2 + 1)
	olcm = lcm(oa, ob)
	if olcm > nlcm:
		print("Nova's gonna kill me")
	else:
		print('YESS(sunglass emo)')