Problem: Try guessing the statement from this picture: 

<image>

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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long i, j, k, m, res = -1, cnt, nt;
long double l, r, lm, rm, n;
long double check(long double num) { return num * (n - num); }
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  ;
  cin >> nt;
  while (nt--) {
    cin >> n;
    if (n == 0) {
      cout << 'Y' << ' ' << fixed << setprecision(10) << 0.0 << ' ' << 0.0
           << endl;
      continue;
    }
    if (n <= 1) {
      cout << 'N' << endl;
      continue;
    }
    l = 1.00000000001;
    r = n - 1.00000000001;
    while (r - l > 0.00000000001) {
      lm = l + (r - l) / 3;
      rm = r - (r - l) / 3;
      if (abs(n - check(lm)) < abs(n - check(rm)))
        r = rm;
      else
        l = lm;
    }
    if (abs((l + (n - l)) - l * (n - l)) < 0.000001 &&
        abs((l + (n - l)) - n) < 0.000001)
      cout << 'Y' << ' ' << fixed << setprecision(10) << l << ' ' << n - l
           << endl;
    else
      cout << 'N' << endl;
  }
  exit(0);
}