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;
const int N = 200500;
double get(double x, double a) {
  double b = x - a;
  return abs(x - b * a);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout << fixed << setprecision(12);
  int tn;
  cin >> tn;
  for (int tc = 0, qwerty = tn; tc < qwerty; tc++) {
    double x;
    cin >> x;
    double l = 0, r = x / 2;
    for (int i = 0, qwerty = 1000; i < qwerty; i++) {
      double a = (l + r) / 2;
      double b = x - a;
      if (a * b > x + 1e-8)
        r = a;
      else
        l = a;
    }
    if (abs((x - l) * l - x) < 1e-7)
      cout << "Y " << x - l << " " << l << "\n";
    else
      cout << "N\n";
  }
}
