#include <bits/stdc++.h>
using namespace std;
long long m, k, t, d, y, z, ans, l, a[300100], b[300100];
string s, f[300100];
long double n, x, w;
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> t;
  cout << fixed << setprecision(9);
  while (t--) {
    cin >> n;
    long double bb = 0, ee = n, mid;
    while (ee - bb > 0.000000000001) {
      mid = (bb + ee) / 2;
      x = n - mid;
      if (x * mid >= n)
        ee = mid;
      else
        bb = mid;
    }
    x = ee * (n - ee);
    y = x;
    if (y == n)
      cout << "Y" << ' ' << n - ee << ' ' << ee << '\n';
    else
      cout << "N" << '\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;
long long m, k, t, d, y, z, ans, l, a[300100], b[300100];
string s, f[300100];
long double n, x, w;
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> t;
  cout << fixed << setprecision(9);
  while (t--) {
    cin >> n;
    long double bb = 0, ee = n, mid;
    while (ee - bb > 0.000000000001) {
      mid = (bb + ee) / 2;
      x = n - mid;
      if (x * mid >= n)
        ee = mid;
      else
        bb = mid;
    }
    x = ee * (n - ee);
    y = x;
    if (y == n)
      cout << "Y" << ' ' << n - ee << ' ' << ee << '\n';
    else
      cout << "N" << '\n';
  }
  return 0;
}

```