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;
struct abc {
  long long int c, x, y;
};
bool compare(abc a1, abc a2) {
  if (a1.x == a2.x) return a1.y > a2.y;
  return a1.x < a2.x;
}
const long long int mod = 1e9 + 7;
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long int t;
  cin >> t;
  while (t--) {
    long double d;
    cin >> d;
    if (d * d - 4 * d < 0)
      cout << "N\n";
    else {
      long double x = (d + sqrt(d * d - 4 * d)) / 2,
                  y = (d - sqrt(d * d - 4 * d)) / 2;
      cout << "Y " << fixed << setprecision(10) << x << ' ' << y << endl;
    }
  }
}
