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;
long long n, x;
int main() {
  cin >> n;
  for (int o = 0; o < n; o++) {
    cin >> x;
    if (x == 0) {
      cout << "Y 0.0 0.0\n";
    } else if (x < 4) {
      cout << "N\n";
    } else if (x == 4) {
      cout << "Y 2.0 2.0\n";
    } else {
      double poc = 1.0;
      double kon = 2.0;
      double num = (double)x;
      for (int i = 1; i <= 1000; i++) {
        double sr = (poc + kon) / 2.0;
        if ((x - sr) * sr > x)
          kon = sr;
        else
          poc = sr;
      }
      printf("Y %.10lf %.10lf\n", x - poc, poc);
    }
  }
  return 0;
}
