#include <bits/stdc++.h>
using namespace std;
void solve() {
  double d;
  cin >> d;
  if (d == 1 || d == 2 || d == 3) {
    cout << "N" << '\n';
  } else {
    cout << "Y"
         << " ";
    if (((d + sqrt((d * d) - 4 * d)) / 2) == 0) {
      cout << 0 << " " << 0 << '\n';
    } else {
      cout << fixed << setprecision(9) << (d + sqrt((d * d) - 4 * d)) / 2 << " "
           << d / ((d + sqrt((d * d) - 4 * d)) / 2) << '\n';
    }
  }
}
int32_t main() {
  long long int t = 1;
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> t;
  while (t--) {
    solve();
  }
}

### Prompt

Create a solution in CPP for 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;
void solve() {
  double d;
  cin >> d;
  if (d == 1 || d == 2 || d == 3) {
    cout << "N" << '\n';
  } else {
    cout << "Y"
         << " ";
    if (((d + sqrt((d * d) - 4 * d)) / 2) == 0) {
      cout << 0 << " " << 0 << '\n';
    } else {
      cout << fixed << setprecision(9) << (d + sqrt((d * d) - 4 * d)) / 2 << " "
           << d / ((d + sqrt((d * d) - 4 * d)) / 2) << '\n';
    }
  }
}
int32_t main() {
  long long int t = 1;
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> t;
  while (t--) {
    solve();
  }
}

```