Problem: Try guessing the statement from this picture: 

<image>

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 long long INF = 1e18;
template <typename T>
void in(std::vector<T> &v) {
  for (auto &now : v) {
    cin >> now;
  }
}
template <typename T>
void in2(std::vector<T> &v) {
  for (auto &now : v) {
    in(now);
  }
}
template <typename T>
void pv(std::vector<T> &v) {
  for (auto now : v) {
    cout << now << " ";
  }
  cout << "\n";
}
template <typename T>
void pv2(std::vector<T> &v) {
  for (auto now : v) {
    pv(now);
  }
  cout << "\n";
}
int main() {
  int t;
  cin >> t;
  for (int i = 0; i < t; ++i) {
    long double d;
    cin >> d;
    if (d == 0) {
      cout << "Y " << 0.0 << " " << 0.0 << "\n";
    } else if (d - 4 < 0) {
      cout << "N\n";
    } else {
      long double a = 0.5 * (d - sqrt((d - 4) * d));
      cout << "Y ";
      cout << setprecision(10) << fixed << a << " " << d - a << "\n";
    }
  }
  return 0;
}