#include <bits/stdc++.h>
using namespace std;
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    double x, y, z, a, b, c, sq;
    cin >> x;
    a = 1.0, b = x, c = x;
    z = (b * b) - (4 * a * c);
    if (z < 0) {
      cout << "N" << endl;
      continue;
    }
    y = ((-b + (sqrt((b * b) - (4 * a * c)))) / 2 * a);
    if (y != 0) y = -1 * y;
    cout << fixed << setprecision(9) << "Y " << (x - y) << " " << y << endl;
  }
}

### Prompt

Please create a solution in Cpp 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 n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    double x, y, z, a, b, c, sq;
    cin >> x;
    a = 1.0, b = x, c = x;
    z = (b * b) - (4 * a * c);
    if (z < 0) {
      cout << "N" << endl;
      continue;
    }
    y = ((-b + (sqrt((b * b) - (4 * a * c)))) / 2 * a);
    if (y != 0) y = -1 * y;
    cout << fixed << setprecision(9) << "Y " << (x - y) << " " << y << endl;
  }
}

```