#include <bits/stdc++.h>
using namespace std;
bool sma(long double a, long double n) {
  if (a * (n - a) < n)
    return true;
  else
    return false;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  for (int i = 0; i < t; i++) {
    int n;
    cin >> n;
    long double l = 0, r = n / 2;
    while (r - l > 1e-12) {
      long double mid = (l + r) / 2;
      if (sma(mid, n))
        l = mid;
      else
        r = mid;
    }
    if (abs(r * (n - r) - n) <= 1e-7)
      cout << "Y " << fixed << setprecision(10) << n - r << " " << r << endl;
    else
      cout << "N" << endl;
  }
  return 0;
}

### Prompt

Your task is to create a CPP solution 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;
bool sma(long double a, long double n) {
  if (a * (n - a) < n)
    return true;
  else
    return false;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  for (int i = 0; i < t; i++) {
    int n;
    cin >> n;
    long double l = 0, r = n / 2;
    while (r - l > 1e-12) {
      long double mid = (l + r) / 2;
      if (sma(mid, n))
        l = mid;
      else
        r = mid;
    }
    if (abs(r * (n - r) - n) <= 1e-7)
      cout << "Y " << fixed << setprecision(10) << n - r << " " << r << endl;
    else
      cout << "N" << endl;
  }
  return 0;
}

```