Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t;
  cin >> t;
  while (t--) {
    double a, b, d;
    cin >> d;
    if (d == 1 || d == 2 || d == 3)
      cout << "N" << endl;
    else if (d == 0) {
      cout << "Y ";
      cout << fixed << setprecision(8) << d << " ";
      cout << fixed << setprecision(8) << d << endl;
    } else {
      b = (d + sqrt(d * (d - 4))) / 2;
      cout << "Y ";
      cout << fixed << setprecision(10) << b << " ";
      cout << fixed << setprecision(10) << (b / (b - 1)) << endl;
    }
  }
  return 0;
}