Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 1;
const double eps = 1e-7;
const int pw = 257;
inline double check(double d, double x) { return (d - x) * x; }
int main() {
  ios_base ::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    double d;
    cin >> d;
    if (d - 0.0 < eps) {
      cout << "Y 0.000000000 0.000000000\n";
      continue;
    }
    if (d - 1.0 < eps) {
      cout << "N\n";
      continue;
    }
    double l = 1.0, r = 2.0;
    for (int i = 1; i <= 10000; ++i) {
      double mid = (l + r) / 2.0;
      if (check(d, mid) < d)
        l = mid;
      else
        r = mid;
    }
    if (-eps < (d - r) * r - d && (d - r) * r - d < eps) {
      cout << "Y ";
      cout << setprecision(13) << fixed << d - r << ' ' << r << "\n";
    } else
      cout << "N\n";
  }
  return 0;
}