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;
struct Testcase {
  double n;
  Testcase(int n) : n(n){};
  pair<double, double> solve(bool &status) {
    pair<double, double> a = {0, n};
    status = true;
    double m = n / 2;
    if (m * m < n) {
      status = false;
      return a;
    }
    if (m * m == n) {
      return {m, m};
    }
    double lo = 0, hi = m;
    for (int i = 0; i < 100; ++i) {
      double mid = (lo + hi) / 2;
      double res = mid * (n - mid);
      if (res == n) {
        return {mid, n - mid};
      } else if (res < n) {
        lo = mid;
      } else if (res > n) {
        hi = mid;
      }
    }
    return {lo, n - lo};
  }
};
int main() {
  int t;
  cin >> t;
  for (int i = 0; i < t; ++i) {
    int n;
    cin >> n;
    bool status = true;
    auto [a, b] = Testcase(n).solve(status);
    if (not status) {
      cout << "N" << endl;
    } else {
      cout << fixed << setprecision(15) << "Y " << a << " " << b << endl;
    }
  }
}