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;
const int maxx = 1e5 + 7;
const int Inf = 1 << 30;
int cas;
int n;
int main() {
  cin >> cas;
  while (cas--) {
    cin >> n;
    int dd = n * (n - 4);
    if (dd < 0)
      puts("N");
    else {
      double t = sqrt(dd);
      double a = (n + t) / 2;
      double b = (n - t) / 2;
      if (a < 0 || b < 0) {
        puts("N");
      } else {
        puts("Y");
        printf("%.9lf %.9lf\n", a, b);
      }
    }
  }
}
