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;
const double pi = 2 * acos(0.0);
double inff = 1.0 / 0.0;
long long int x, y, n, i, z, j, k, w, u, q, c, t, m;
double func(double d, double a) { return (a * a - a * d + d); }
void solve() {
  double d;
  cin >> d;
  cout << setprecision(9) << fixed;
  if (d > 0 and d < 4) {
    cout << "N"
         << "\n";
    return;
  }
  cout << "Y ";
  double lo = 0, hi = d, mid;
  while (hi > lo + (1e-14) * lo) {
    mid = (hi + lo) / 2;
    if ((func(d, mid) > 0 and mid < d / 2) or
        (func(d, mid) < 0 and mid >= d / 2))
      lo = mid + (1e-14);
    else
      hi = mid;
  }
  cout << lo << " " << (d - lo) << "\n";
}
int main(void) {
  ios::sync_with_stdio(false), cout.tie(0), cin.tie(0);
  ;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}
