Try guessing the statement from this picture: 

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You are given a non-negative integer d. You have to find two non-negative real numbers a and b such that a + b = d and a ⋅ b = d.

Input

The first line contains t (1 ≤ t ≤ 10^3) — the number of test cases.

Each test case contains one integer d (0 ≤ d ≤ 10^3).

Output

For each test print one line.

If there is an answer for the i-th test, print "Y", and then the numbers a and b.

If there is no answer for the i-th test, print "N".

Your answer will be considered correct if |(a + b) - a ⋅ b| ≤ 10^{-6} and |(a + b) - d| ≤ 10^{-6}.

Example

Input


7
69
0
1
4
5
999
1000


Output


Y 67.985071301 1.014928699
Y 0.000000000 0.000000000
N
Y 2.000000000 2.000000000
Y 3.618033989 1.381966011
Y 997.998996990 1.001003010
Y 998.998997995 1.001002005
t = int(input())
import math

def p(x):
	y = list(str(x))
	flag=1
	z = y.index('.')
	for i in range(min(len(y),z+1+9)):
		if min(len(y),z+1+9)==(z+1+9):
			flag = 0
		print(y[i],end="")
	if flag==1:
		printed = len(y) - z - 1
		for i in range(9-printed):
			print("0",end="")
	print(end=" ")

for i in range(t):
	d = int(input())
	a1 = 0
	a2 = 0
	try:
		a1 = (d+math.sqrt(d**2-4*d))/2
		a2 = (d-math.sqrt(d**2-4*d))/2
	except:
		print("N")
		continue	

	if d==0:
		print("Y 0.000000000 0.000000000",end="")
	# if d==1:
	# 	print("N",end=" ")
	else:
		if a1>0:
			print("Y",end=" ")
			p(a1)
			p(d/a1)
		elif a2>0:
			print("Y",end=" ")
			p(a2)
			p(d/a2)
		else:
			print("N")
	print()

