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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.io.*;
import java.util.*;

public class Main {

    public static void main(String[] args) throws IOException {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        for (int i = 0; i < t; i++) {
            int d = sc.nextInt();
            int delta = d * d - 4 * d;
            if (delta < 0) {
                System.out.println("N");
                continue;
            }
            double sqdelta = Math.sqrt(delta);
            double b = (d + sqdelta) / 2;
            double a = d - b;
            if (a < 0 || b < 0) {
                System.out.println("N");
            } else {
                System.out.println("Y " + a + " " + b);
            }
        }
    }

    static class Scanner {
        StringTokenizer st;
        BufferedReader br;
        public Scanner(InputStream s) {
            br = new BufferedReader(new InputStreamReader(s));
        }
        public String next() throws IOException {
            while (st == null || !st.hasMoreTokens()) {
                st = new StringTokenizer(br.readLine());
            }
            return st.nextToken();
        }
        public int nextInt() throws IOException {
            return Integer.parseInt(next());
        }
        public long nextLong() throws IOException {
            return Long.parseLong(next());
        }
    }
}
