#include <bits/stdc++.h>
using namespace std;
const double PI = 3.141592653589793238460;
const long long mod = 1e9 + 7;
const long long INF = (long long)1e18;
void FAST() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
}
void solve() {
  long long d;
  cin >> d;
  if (d > 0 && d < 4) {
    cout << "N\n";
    return;
  }
  double D = sqrt(d * d - 4 * d);
  double a = (d + D) / 2;
  double b = (d - D) / 2;
  cout << "Y " << fixed << setprecision(12) << a << " " << b << '\n';
}
int32_t main() {
  FAST();
  long long t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  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;
const double PI = 3.141592653589793238460;
const long long mod = 1e9 + 7;
const long long INF = (long long)1e18;
void FAST() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
}
void solve() {
  long long d;
  cin >> d;
  if (d > 0 && d < 4) {
    cout << "N\n";
    return;
  }
  double D = sqrt(d * d - 4 * d);
  double a = (d + D) / 2;
  double b = (d - D) / 2;
  cout << "Y " << fixed << setprecision(12) << a << " " << b << '\n';
}
int32_t main() {
  FAST();
  long long t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

```