#include <bits/stdc++.h>
using namespace std;
int main() {
  int t, res, i, j, k, x;
  long double a, b, d;
  cin >> t;
  for (i = 1; i <= t; i++) {
    cin >> d;
    res = 0;
    if ((d - 0) <= 0.00000001) {
      a = 0.0000000;
      b = 0.0000000;
    } else if (d < 4)
      res = -1;
    else {
      a = d + sqrt(d * d - 4 * d);
      a = a / 2;
      b = d - a;
    }
    if (res == 0) {
      cout << "Y ";
      cout << fixed << showpoint << setprecision(9) << a << " ";
      cout << fixed << showpoint << setprecision(9) << b << endl;
    } else
      cout << "N" << endl;
  }
  return 0;
}

### Prompt

Your challenge is to write a CPP solution to the following 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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
  int t, res, i, j, k, x;
  long double a, b, d;
  cin >> t;
  for (i = 1; i <= t; i++) {
    cin >> d;
    res = 0;
    if ((d - 0) <= 0.00000001) {
      a = 0.0000000;
      b = 0.0000000;
    } else if (d < 4)
      res = -1;
    else {
      a = d + sqrt(d * d - 4 * d);
      a = a / 2;
      b = d - a;
    }
    if (res == 0) {
      cout << "Y ";
      cout << fixed << showpoint << setprecision(9) << a << " ";
      cout << fixed << showpoint << setprecision(9) << b << endl;
    } else
      cout << "N" << endl;
  }
  return 0;
}

```