Problem: Try guessing the statement from this picture: 

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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;
const long long inf = 1e18;
const long long N = 200005;
long long n;
void solve() {
  long long i, j, k;
  long long d;
  cin >> d;
  cout << fixed << setprecision(9);
  if (d == 0) {
    cout << "Y " << 0.0 << " " << 0.0 << "\n";
    return;
  }
  if (d < 4) {
    cout << "N"
         << "\n";
    return;
  }
  double D = sqrt(d * (d - 4));
  double a = (d + D) / 2.0;
  double b = (d - D) / 2.0;
  cout << "Y " << a << " " << b << "\n";
  return;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long t = 1, i, j, k;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}