#include <bits/stdc++.h>
using namespace std;
int main() {
  cout.precision(15);
  cout.setf(ios::floatfield, ios::fixed);
  long t;
  cin >> t;
  while (t--) {
    double a1, a2, b1, b2, c, r, d;
    cin >> d;
    r = d * d - 4 * d;
    if (r < 0) {
      cout << "N" << endl;
      continue;
    }
    a1 = (d + sqrt(r)) / 2;
    b1 = d - a1;
    if (a1 >= 0 && b1 >= 0) {
      cout << "Y"
           << " " << a1 << " " << b1 << endl;
      continue;
    }
    a2 = (d - sqrt(r)) / 2;
    b2 = d - a2;
    if (a2 >= 0 && b2 >= 0) {
      cout << "Y"
           << " " << a2 << " " << b2 << endl;
      continue;
    }
    cout << "N" << endl;
  }
}

### Prompt

Develop a solution in cpp to the problem described below:

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() {
  cout.precision(15);
  cout.setf(ios::floatfield, ios::fixed);
  long t;
  cin >> t;
  while (t--) {
    double a1, a2, b1, b2, c, r, d;
    cin >> d;
    r = d * d - 4 * d;
    if (r < 0) {
      cout << "N" << endl;
      continue;
    }
    a1 = (d + sqrt(r)) / 2;
    b1 = d - a1;
    if (a1 >= 0 && b1 >= 0) {
      cout << "Y"
           << " " << a1 << " " << b1 << endl;
      continue;
    }
    a2 = (d - sqrt(r)) / 2;
    b2 = d - a2;
    if (a2 >= 0 && b2 >= 0) {
      cout << "Y"
           << " " << a2 << " " << b2 << endl;
      continue;
    }
    cout << "N" << endl;
  }
}

```