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
import sys
import itertools
import math
import collections
from collections import Counter

#########################
# imgur.com/Pkt7iIf.png #
#########################

def sieve(n):
    prime = [True for i in range(n + 1)]
    p = 2
    while (p * p <= n):
        if (prime[p] == True):
            for i in range(p * 2, n + 1, p):
                prime[i] = False
        p += 1
    prime[0] = prime[1] = False
    r = [p for p in range(n + 1) if prime[p]]
    return r
def divs(n, start=1):
    r = []
    for i in range(start, int(math.sqrt(n) + 1)):
        if (n % i == 0):
            if (n / i == i):
                r.append(i)
            else:
                r.extend([i, n // i])
    return r
def ceil(n, k): return n // k + (n % k != 0)
def ii(): return int(input())
def mi(): return map(int, input().split())
def li(): return list(map(int, input().split()))
def lcm(a, b): return abs(a * b) // math.gcd(a, b)
def prr(a, sep=' '): print(sep.join(map(str, a)))
def dd(): return collections.defaultdict(int)

e = 10**(-6)

def solve(n, l, r):

    for i in range(500):
        a = l + (r - l) / 2
        b = n - a
        #print(a * b, a + b)
        if abs(a + b - a * b) <= e and abs(a + b - n) <= e:
            return a
        if a * b - n > e:
            r = a - e
        else:
            l = a + e
    return -1

t = ii()
for _ in range(t):
    n = ii()
    a = solve(n, 0, n)
    print('Y', a, n - a) if a != -1 else print('N')












