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;
const int infint = 1e9;
const int MOD = (int)1e9 + 7;
const int MAXN = (int)1e5 + 7;
long long T;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> T;
  for (int i = 0; i < T; i++) {
    long double d;
    cin >> d;
    long double a = -1, b = d, c = -d;
    long double delta = b * b - 4 * a * c;
    if (delta < 0) {
      cout << "N\n";
      continue;
    }
    long double A1 = (-b + sqrt(delta)) / (2 * a);
    long double A2 = (-b - sqrt(delta)) / (2 * a);
    cout << "Y ";
    cout << fixed << setprecision(9) << A1 << " " << A2 << "\n";
  }
}
