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 long double EPS = 1e-16;
int main() {
  std::ios_base::sync_with_stdio(0);
  int cases;
  long double d;
  cin >> cases;
  while (cases--) {
    cin >> d;
    long double low = 0.0, high = d * 0.5, mid, x, y;
    for (int i = 1; i <= 100; i++) {
      mid = (low + high) * 0.5;
      x = mid, y = d - mid;
      if (x * y < d || abs(x * y - d) < EPS)
        low = mid;
      else
        high = mid;
    }
    x = mid, y = d - mid;
    if (abs(x * y - d) < 0.0000001)
      cout << fixed << setprecision(12) << "Y " << x << " " << y << '\n';
    else
      cout << "N" << '\n';
  }
  return 0;
}
