#include <bits/stdc++.h>
using namespace std;
using ll = long long int;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using maxHeap = priority_queue<int>;
using minHeap = priority_queue<int, vector<int>, greater<int>>;
int main() {
  ios::sync_with_stdio(0), cin.tie(0);
  int t, x;
  double d, p;
  cout << fixed << setprecision(10);
  for (cin >> t; t; t--) {
    cin >> x;
    d = x;
    if (x * x - 4 * x < 0) {
      cout << "N\n";
    } else {
      p = sqrt(d * d - 4 * d);
      cout << "Y " << (d - p) / 2 << ' ' << (d + p) / 2 << '\n';
    }
  }
  return 0;
}

### 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;
using ll = long long int;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using maxHeap = priority_queue<int>;
using minHeap = priority_queue<int, vector<int>, greater<int>>;
int main() {
  ios::sync_with_stdio(0), cin.tie(0);
  int t, x;
  double d, p;
  cout << fixed << setprecision(10);
  for (cin >> t; t; t--) {
    cin >> x;
    d = x;
    if (x * x - 4 * x < 0) {
      cout << "N\n";
    } else {
      p = sqrt(d * d - 4 * d);
      cout << "Y " << (d - p) / 2 << ' ' << (d + p) / 2 << '\n';
    }
  }
  return 0;
}

```