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;
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    double b1, b2, d;
    cin >> d;
    b1 = (-d + sqrt(d * d - ((-4) * (-d)))) / (-2);
    b2 = (-d - sqrt(d * d - ((-4) * (-d)))) / (-2);
    double a1, a2;
    a1 = d - b1;
    a2 = d - b2;
    if (d == 0) {
      puts("Y 0.000000000 0.000000000");
      continue;
    }
    cout << fixed;
    if (a1 > -1e9 && b1 > -1e9 && fabs((a1 + b1) - (a1 * b1)) <= 1e6 &&
        fabs((a1 + b1) - d) <= 1e6) {
      cout << "Y ";
      cout << setprecision(9) << a1;
      cout << " ";
      cout << setprecision(9) << b1;
      cout << '\n';
    } else if (a2 > -1e9 && b2 > -1e9 && fabs((a2 + b2) - (a2 * b2)) <= 1e6 &&
               fabs((a2 + b2) - d) <= 1e6) {
      cout << "Y ";
      cout << setprecision(9) << a2;
      cout << " ";
      cout << setprecision(9) << b2;
      cout << '\n';
    } else {
      puts("N");
    }
  }
  return 0;
}
