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 double pi = 3.14159265358979323846;
void rw() {}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long q;
  cin >> q;
  while (q--) {
    long double a, b, c, t = 2;
    long long n;
    cin >> n;
    if (n > 3) {
      c = sqrt(n * n - 4 * n);
      a = (n - c) / t;
      b = (n + c) / t;
      cout << "Y " << setprecision(9) << fixed << b << " " << a;
      goto done;
    } else if (!n) {
      cout << "Y 0 0";
      goto done;
    }
    cout << "N";
  done:;
    cout << "\n";
  }
  return 0;
}