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 JAVA):

Solution:
import java.util.Scanner;

public class Main {
    public static void main (String[] args) throws java.lang.Exception {
        try {
            Scanner sc=new Scanner(System.in);
            int d=sc.nextInt();
            double r1,r2;
            while (d--!=0) {
                int t = sc.nextInt();
                double del = Math.sqrt((Math.pow(t, 2)) - (4 * t));
                if (t==0){
                    System.out.println("Y 0.000000000000 0.000000000000");
                }
                else if (t<4){
                    System.out.println("N");
                }
                else {
                    if (del < 0) {
                        System.out.println("N");
//                    } else if (del == 0) {
//                        r1 = 0;
//                        r2 = 0;
//                        System.out.println("Y" + " " + r1 + " " + r2);
                    }
                        else {
                        r1 = (t + del) / 2;
                        r2 = (t - del) / 2;
                        System.out.println("Y" + " " + r1 + " " + r2);
                    }
                }
            }
            


        } catch(Exception e) {
            return;
        }
    }

}