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
def bin_search(number):
    left_bound = 0
    right_bound = number
    middle = right_bound + left_bound / 2
    while (abs(number - middle * (number - middle)) >= 0.000001):
        middle = (right_bound + left_bound) / 2
        if middle * (number - middle) > number:
            right_bound = middle
        else:
            left_bound = middle
    return middle, number - middle

amount = int(input())
#print(bin_search(1))
for i in range(amount):
    number = int(input())
    if number < 4 and number > 0:
        print('N')
    else:
        first, second = bin_search(number)
        #print( abs(first + second - first * second), abs(first + second - number))
        if abs(first + second - first * second) <= 10e-6 and abs(first + second - number) <= 10e-6:
            print('Y', first, second)
        else:
            print('difference')
        #print("Y", first, second)
    
"""a + b = d
a * b = d
=> a = d - b
=> (d - b) * b = d
=> db - b^2 = d
=> -b^2 - db - d = 0
=> b^2 + db + d = 0
D = d^2 - 4d
d(d - 4) < 0 => d (0;4) doesn't have any solutions"""