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
from __future__ import division
from sys import stdin, stdout


def write(x):
    stdout.write(str(x) + "\n")

def solve(d):
    if d == 0:
        return 0., 0.
    if d < 4:
        return None

    high = 2
    low = 1

    while high - low > 0.00000000001:
        mid = (high + low) / 2

        if (d - mid) * mid > d:
            high = mid
        else:
            low = mid

    return d - mid, mid

t = int(stdin.readline())
for line in stdin.readlines():
    d = int(line)
    res = solve(d)
    if res:
        write("Y {} {}".format(res[0], res[1]))
    else:
        write("N")
