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;
void solve() {
  double n;
  cin >> n;
  if (n == 1)
    cout << "N\n";
  else if (n == 0)
    cout << "Y 0 0\n";
  else {
    double x = -(2 - n);
    if ((n * n) - (4 * n) < 0)
      cout << "N\n";
    else {
      double dispo = sqrt(((n * n) - (4 * n)));
      double root1 = x - dispo, root2 = x + dispo;
      root1 /= 2.0;
      double f = 1 + root1;
      cout << "Y " << (f) << " " << n - (f) << "\n";
    }
  }
}
int main() {
  cout << fixed << setprecision(13);
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}