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
import java.util.*;
import static java.lang.Math.*;
import java.io.*;

public class SolutionB {
      public static void main(String args[])throws IOException{
              Scanner sc = new Scanner(System.in);
              int t = sc.nextInt();
              while(t > 0){
                     t--;
                     int n = sc.nextInt();
                     double a = 0;
                     double b = 0;
                     if(n == 0){
                            System.out.println("Y " + a + " " + b);
                            continue;
                     }else if(n < 4){
                            System.out.println("N");
                            continue;
                     }
                     double l = 0;
                     double h = n;
                     double e1 = 0;
                     double e2 = 0;
                     double mid = 0;
                     do{
                            mid = (l+h)/2d;
                            //n = a + n/a;
                            double nn = mid + (double)n/mid;
                            if(nn > n){
                                   h = mid;
                            }else{
                                   l = mid;
                            }
                            b = n/mid;
                            e1 = Math.abs((mid + b) - mid * b);
                            e2 = Math.abs((mid + b) - (double)n);
                     }while(e1 > 0.000001 || e2 > 0.000001);
                     //double b = n/l;
                     System.out.println("Y " + mid + " " + b);
              }
      }
}