#include <bits/stdc++.h>
using namespace std;
const int N = 3 * 1e5 + 5;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  double a[1005], b[1005];
  for (int i = 1; i <= t; ++i) {
    int d;
    cin >> d;
    if (d < 4 && d > 0) {
      a[i] = -1;
      b[i] = -1;
      continue;
    }
    double discr = sqrt((d * d - 4 * d) * 1.0);
    a[i] = (d + discr) / 2;
    b[i] = (d - discr) / 2;
  }
  for (int i = 1; i <= t; ++i) {
    if (a[i] == -1)
      cout << "N\n";
    else {
      cout << "Y ";
      cout << fixed << setprecision(9) << a[i] << ' ' << b[i] << '\n';
    }
  }
  return 0;
}

### Prompt

In cpp, your task is to solve 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;
const int N = 3 * 1e5 + 5;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  double a[1005], b[1005];
  for (int i = 1; i <= t; ++i) {
    int d;
    cin >> d;
    if (d < 4 && d > 0) {
      a[i] = -1;
      b[i] = -1;
      continue;
    }
    double discr = sqrt((d * d - 4 * d) * 1.0);
    a[i] = (d + discr) / 2;
    b[i] = (d - discr) / 2;
  }
  for (int i = 1; i <= t; ++i) {
    if (a[i] == -1)
      cout << "N\n";
    else {
      cout << "Y ";
      cout << fixed << setprecision(9) << a[i] << ' ' << b[i] << '\n';
    }
  }
  return 0;
}

```