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;
vector<long long> primes(1, 2);
bool isPrime(long long n) {
  if (n == 2) return true;
  for (int i = 2; i * i <= n; ++i)
    if (n % i == 0) return false;
  return true;
}
long long smallestPrimeDiv(long long n) {
  for (int i = 0; i < primes.size(); ++i) {
    if (n % primes[i] == 0) return primes[i];
  }
  int num = primes[primes.size() - 1] + 2;
  while (1) {
    if (isPrime(num)) {
      primes.push_back(num);
      if (n % num == 0) return num;
    }
    num += 2;
  }
}
int main() {
  int n;
  cin >> n;
  vector<double> v(n);
  for (int i = 0; i < n; ++i) {
    cin >> v[i];
  }
  for (int i = 0; i < n; ++i) {
    double d = v[i];
    if (d == 0) {
      cout << setprecision(11);
      cout << "Y " << 0.000000 << " " << 0.000000 << endl;
    } else if (d < 4) {
      cout << "N" << endl;
    } else {
      double b = ((d + sqrt(d * d - 4 * d)) / 2);
      double a = d - b;
      cout << setprecision(11);
      cout << "Y " << b << " " << a << endl;
    }
  }
}