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
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e5 + 5;
vector<int> prim;
bool used[MAXN];
int main() {
  cin.tie(0);
  cin.sync_with_stdio(0);
  double d, a, b;
  int n;
  cout.precision(15);
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> d;
    double dd = d * d - 4 * d;
    if (dd < 0) {
      cout << "N" << endl;
    } else {
      double b1, b2;
      b1 = (-d + sqrt(dd)) / (-2);
      b2 = (-d - sqrt(dd)) / (-2);
      if ((b1 < 0) || (b1 > d)) {
        if ((b2 < 0) || (b2 > d)) {
          cout << "N" << endl;
        } else {
          cout << fixed << "Y"
               << " " << b2 << " " << d - b2 << endl;
        }
      } else {
        cout << fixed << "Y"
             << " " << b2 << " " << d - b2 << endl;
      }
    }
  }
  return (0);
}
