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 double eps = 1e-10;
long long My_Pow(long long a, long long k) {
  long long res = 1;
  for (int i = 0; i < k; i++) {
    res *= a;
  }
  return res;
}
int My_Pow(int a, int k) {
  int res = 1;
  for (int i = 0; i < k; i++) {
    res *= a;
  }
  return res;
}
long long gcd(long long a, long long b) {
  while (b) {
    a %= b;
    swap(a, b);
  }
  return a;
}
int gcd(int a, int b) {
  while (b) {
    a %= b;
    swap(a, b);
  }
  return a;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cout << fixed << setprecision(10);
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int q;
  cin >> q;
  while (q--) {
    long double d;
    cin >> d;
    long double l = 0;
    long double r = d;
    for (int i = 0; i < 1000; i++) {
      long double a = (l + r) / 2.0;
      long double b = d - a;
      if (a * b > d - eps) {
        l = a;
      } else {
        r = a;
      }
    }
    if (fabs(l * (d - l) - d) < eps) {
      cout << "Y " << l << " " << d - l << "\n";
    } else {
      cout << "N"
           << "\n";
    }
  }
  return 0;
}