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
#include <bits/stdc++.h>
using namespace std;
ostream &operator<<(ostream &out, vector<long long int> p) {
  long long int n = p.size();
  for (long long int i = (long long int)(0); i < (long long int)(n); i++) {
    if (i) cout << endl;
    out << p[i];
  }
  out << endl;
  return out;
}
int main() {
  long long int n2, n;
  cin >> n2;
  while (n2--) {
    cin >> n;
    if ((n > 0) && (n < 4)) {
      cout << "N" << endl;
      continue;
    }
    cout << "Y ";
    long double rpta = (sqrt(n * n - 4 * n) + n * 1.0) / 2.0;
    cout.precision(9);
    cout << fixed;
    cout << rpta << " " << n - rpta << endl;
  }
  return 0;
}
