Problem: Try guessing the statement from this picture: 

<image>

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 double pp(long double n) {
  long double l = 1;
  long double r = 2;
  while (abs(l - r) >= 1e-12) {
    long double m = (l + r) / 2;
    if ((n - m) - (n / m) > 0)
      r = m;
    else
      l = m;
  }
  return r;
}
int main() {
  int q;
  cin >> q;
  long double n;
  for (int i = 0; i < q; i++) {
    cin >> n;
    if (n == 0) {
      cout << "Y 0.000000000 0.000000000" << endl;
    } else if (n == 1 || n == 2 || n == 3) {
      cout << "N" << endl;
    } else {
      cout << fixed << setprecision(9) << "Y " << n - pp(n) << " " << pp(n)
           << endl;
    }
  }
}