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
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
using namespace std;
const int INF = 1e9 + 5;
const int MOD = 1e9 + 7;
double pi = 2 * acos(0.0);
void solve() {
  int d;
  cin >> d;
  if (d == 0) {
    cout << "Y 0 0\n";
    return;
  }
  int D = d * d - (4 * d);
  if (d > 0 and d < 4) {
    cout << "N\n";
    return;
  }
  double a = ((double)d + (double)(sqrt(D))) / 2.0,
         b = ((double)d - (double)(sqrt(D))) / 2.0;
  cout << "Y ";
  cout << fixed << setprecision(20) << a << ' ' << b << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  int t = 1;
  cin >> t;
  while (t--) solve();
  return 0;
}
