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;
double bin(double l, double r, double d) {
  if (r - l < 0.00000000001) return l;
  double mid = (r + l) / 2.0;
  if (mid * (d - mid) > d) {
    l = mid;
  } else {
    r = mid;
  }
  return bin(l, r, d);
}
int main() {
  int n;
  double ans, d;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> d;
    if (d <= 3 and d != 0)
      cout << "N" << endl;
    else if (d == 0)
      cout << "Y" << ' ' << "0.000000000" << ' ' << "0.000000000" << endl;
    else {
      ans = bin(d - 2, d - 1, d);
      cout << "Y" << std::fixed << setprecision(12) << ' ' << ans << ' '
           << d - ans << endl;
    }
  }
  return 0;
}
