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;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int d;
    cin >> d;
    long double a, b;
    if (d != 0 && d < 4) {
      printf("N\n");
    } else {
      long double e = d;
      if (d == 0) {
        a = 0;
        b = 0;
      } else if (d == 4) {
        a = 2;
        b = 2;
      } else {
        long double delta = 1;
        a = 1;
        b = d - 1;
        for (int i = 0; i < 50; i++) {
          while (a * b < d) {
            a += delta;
            b -= delta;
          }
          a -= delta;
          b += delta;
          delta /= 2;
        }
      }
      printf("Y %0.9Lf %0.9Lf\n", a, b);
    }
  }
  return 0;
}
