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 run() {
  double d;
  cin >> d;
  double delta = d * d - 4 * d;
  if (delta < 0) {
    printf("N");
    return 0;
  }
  double b1 = (+d - sqrt(delta)) / 2.0;
  double b2 = (+d + sqrt(delta)) / 2.0;
  double a1 = d - b1;
  double a2 = d - b2;
  if (a1 >= 0.0 && b1 >= 0.0) {
    printf("Y %0.9f %0.9f", a1, b1);
    return 0;
  }
  if (a2 >= 0.0 && b2 >= 0.0) {
    printf("Y %0.9f %0.9f", a2, b2);
    return 0;
  }
  printf("N");
}
int main() {
  int t;
  cin >> t;
  while (t--) run(), printf("\n");
  return 0;
}
