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
#include <bits/stdc++.h>
using namespace std;
void solve() {
  int d;
  cin >> d;
  if (d == 0) {
    cout << 'Y' << ' ' << 0.0 << ' ' << 0.0 << endl;
    return;
  }
  if (d < 4) {
    cout << 'N' << endl;
    return;
  }
  double bound = 0.000000001;
  double l = 2.0;
  double r = d - 1;
  while (r - l > bound) {
    double mid = (r + l) / 2;
    if (mid * (d - mid) > d) {
      l = mid;
    } else {
      r = mid;
    }
  }
  cout << 'Y' << ' ' << l << ' ' << d - l << endl;
}
int main() {
  cout << setprecision(15) << fixed;
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
}
