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
#include <bits/stdc++.h>
using namespace std;
void tim(long long s, long long p) {
  double d = s * s - 4 * p;
  cout << "Y"
       << " ";
  cout << setprecision(9) << fixed << (s * 1.0 + (double)sqrt(d)) / (2 * 1.0)
       << " ";
  cout << setprecision(9) << fixed << (s * 1.0 - (double)sqrt(d)) / (2 * 1.0)
       << endl;
}
int main() {
  long long n;
  cin >> n;
  long long a[n];
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
  }
  for (long long z = 1; z <= n; z++) {
    if (a[z] * a[z] - 4 * a[z] < 0)
      cout << "N" << endl;
    else {
      long long s = a[z];
      long long p = a[z];
      tim(s, p);
    }
  }
}
