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>
using namespace std;
const long long fn = 1e10;
bool f(double mid, double d) { return (mid * mid - d * mid + d) > 0; }
void solve(double d) {
  if (d * d - 4 * d < 0) {
    cout << "N"
         << "\n";
    return;
  }
  double lo = 0, hi = d / 2 + 1e-15;
  while (abs(lo - hi) > 1e-15) {
    double mid = 0.5 * (lo + hi);
    (f(mid, d) ? lo : hi) = mid;
  }
  if (lo >= 0 && d - lo >= 0) {
    cout << "Y"
         << " " << d - lo << " " << lo << "\n";
    return;
  }
}
int main() {
  int t;
  cin >> t;
  cout << fixed << setprecision(9);
  for (int i = 0; i < t; ++i) {
    double d;
    cin >> d;
    solve(d);
  }
}
