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;
long long inp;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.precision(20);
  long long t;
  cin >> t;
  vector<double> sb, sa;
  double d;
  for (int i = (0); i < (t); i++) {
    cin >> d;
    if (d == 0) {
      sa.push_back(0);
      sb.push_back(0);
    } else if (d < 4) {
      sa.push_back(-1);
      sb.push_back(-1);
    } else {
      double b = (d + sqrt((double)d * d - (double)4 * d)) / 2;
      sb.push_back(b);
      sa.push_back(b / (b - 1));
    }
  }
  for (int i = (0); i < (t); i++) {
    if (sa[i] == -1) {
      cout << "N" << endl;
    } else {
      cout << "Y " << sb[i] << " " << sa[i] << endl;
    }
  }
  return 0;
}