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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const ll MAX = 10000000000000;
const int N = 200005;
const int mod = 1e9 + 7;
void solve() {
  double d;
  cin >> d;
  double a, b;
  a = (d + sqrt(d * d - 4 * d)) / 2;
  b = (d - sqrt(d * d - 4 * d)) / 2;
  if (a >= 0 && b >= 0) {
    cout << fixed << setprecision(9) << "Y " << a << " " << b << endl;
  } else
    cout << "N\n";
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  cin >> t;
  while (t--) solve();
  return 0;
}