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>
const int inf = 1e9 + 69;
const int MAX = 2e3;
using namespace std;
bool compare(double a, double b) { return fabs(a - b) < 0.000001; }
int main() {
  cout.precision(15);
  int nTest;
  cin >> nTest;
  while (nTest--) {
    double n;
    cin >> n;
    double denta = n * n - 4 * n;
    double can = sqrt(denta);
    if (can < 0) {
      cout << "N" << '\n';
    } else {
      double x1 = (n + can) / 2;
      double x2 = (n - can) / 2;
      double y1 = n - x1;
      double y2 = n - x2;
      if (compare(x1 + y1, x1 * y1) and compare(x1 + y1, n)) {
        cout << "Y " << x1 << " " << y1 << '\n';
      } else if (compare(x2 + y2, x2 * y2) and compare(x2 + y2, n)) {
        cout << "Y " << x2 << " " << y2 << '\n';
      } else {
        cout << "N" << '\n';
      }
    }
  }
  return 0;
}
