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 MAXN = 100007;
const double eps = 1e-6;
int main() {
  int n;
  double d;
  cin >> n;
  while (n--) {
    cin >> d;
    double der = sqrt(d * d - 4 * d);
    if (der < 0) {
      cout << "N" << endl;
      continue;
    }
    double cc = (-d - der) / (-2), dd = (-d + der) / (-2);
    if (abs(((d - cc) * cc) - d) < eps)
      printf("Y %.9lf %.9lf\n", cc, d - cc);
    else if (abs(((d - dd) * dd) - d) < eps)
      printf("Y %.9lf %.9lf\n", dd, d - dd);
    else
      cout << "N" << endl;
  }
  return 0;
}
