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
#include <bits/stdc++.h>
using namespace std;
void run_case() {
  int d;
  cin >> d;
  if (d == 0)
    cout << "Y 0.0 0.0\n";
  else if (d < 4)
    cout << "N\n";
  else {
    double delta = sqrt(d * (d - 4));
    double a = (d + delta) / 2.0;
    double b = (d - delta) / 2.0;
    cout << "Y " << a << " " << b << "\n";
  }
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0);
  cout.flags(ios::fixed);
  cout.precision(9);
  int t;
  cin >> t;
  while (t--) run_case();
  cout.flush();
  return 0;
}
