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 main() {
  int n;
  cin >> n;
  double *arr1 = new double[n];
  double d, a1, a2, b1, b2;
  double *arr2 = new double[n];
  char *arr = new char[n];
  for (int i = 0; i < n; i++) {
    cin >> d;
    while (d < 0) cin >> d;
    if (d == 0) {
      arr[i] = 'Y';
      arr1[i] = 0.000000000;
      arr2[i] = 0.000000000;
    } else if (d > 0 && d < 4) {
      arr[i] = 'N';
      arr1[i] = 0.000000000;
      arr2[i] = 0.000000000;
    } else if (d == 4) {
      arr[i] = 'Y';
      arr1[i] = 2.00000000;
      arr2[i] = 2.000000000;
    } else {
      b1 = (d - sqrt(d * d - 4 * d)) / 2;
      a1 = d - b1;
      if (b1 + a1 == d && b1 * a1 == d) {
        arr[i] = 'Y';
        arr1[i] = a1;
        arr2[i] = b1;
      } else {
        b2 = (d + sqrt(d * d - 4 * d)) / 2;
        a2 = d - b2;
        arr[i] = 'Y';
        arr1[i] = a2;
        arr2[i] = b2;
      }
    }
  }
  for (int i = 0; i < n; i++) {
    if (arr[i] == 'N')
      cout << setprecision(11) << arr[i] << endl;
    else
      cout << setprecision(11) << arr[i] << " " << setprecision(11) << arr2[i]
           << " " << setprecision(11) << arr1[i] << endl;
  }
  delete[] arr;
  delete[] arr1;
  delete[] arr2;
  return 0;
}
