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;
int main() {
  ios_base::sync_with_stdio(0), cin.tie(NULL), cout.tie(NULL);
  int t, i, j;
  double n, a, b, lo, hi, mid, mul;
  cin >> t;
  for (i = 0; i < t; i++) {
    cin >> n;
    lo = 0.0;
    hi = n;
    for (j = 0; j <= 100; j++) {
      mid = (lo + hi) / 2.0;
      a = mid;
      b = n - mid;
      mul = a * b;
      if (mul - n > 1e-10) {
        hi = mid;
      } else if (n - mul > 1e-10) {
        lo = mid;
      }
    }
    if (abs(mid * (n - mid) - n) <= 1e-10) {
      cout << "Y " << std::setprecision(12) << mid << ' '
           << std::setprecision(12) << n - mid << endl;
    } else
      cout << "N" << endl;
  }
  return 0;
}
