Try guessing the statement from this picture:  $3$ 

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 $a \cdot b = d$.


-----Input-----

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

Each test case contains one integer $d$ $(0 \le d \le 10^3)$.


-----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 $|(a + b) - a \cdot b| \le 10^{-6}$ and $|(a + b) - d| \le 10^{-6}$.


-----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

Difficulty: interview

Starter code:

Solutions:
for _ in range(int(input())):
    d=int(input())
    anws=False
    if d**2>=4*d:
        root=(d**2-4*d)**0.5
        a=(d+root)/2
        b=(d-root)/2
        anws=True
    if anws:
        print("Y {:.9f} {:.9f}".format(a,b))
    else:
        print("N")

n=int(input())

for i in range(n):
    d=int(input())
    #b**2-bd+d=0
    D=d**2-4*d
    if D>=0:
        b1=(d+D**0.5)/2
        b2=(d-D**0.5)/2
    if D<0 or (b1<0 and b2<0):
        print("N")
    else:
        a1=d-b1
        a2=d-b2
        if a1>=0 and b1>=0:
            print("Y", "%.9f"%a1, "%.9f"%b1)
        elif a2>=0 and b2>=0:
            print("Y", "%.9f"%a2, "%.9f"%b2)
        else:
            print("N")