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;
double a, b, r, s, d;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    cin >> d;
    a = 0, b = d;
    r = 1, s = 0;
    while (r >= 0.0000000000001) {
      s = 0;
      double x = a, y = b;
      while (s < 10 * r) {
        if ((a + s) * (b - s) > x * y and (a + s) * (b - s) <= d) {
          x = a + s;
          y = b - s;
        }
        s += r;
      }
      a = x, b = y;
      r /= 10;
    }
    double diff = d - a * b;
    if (diff > 0.000001) {
      cout << "N\n";
    } else {
      cout << "Y ";
      cout << setprecision(11) << a << " " << b << "\n";
    }
  }
  return 0;
}
