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;
void solve() {
  double d;
  scanf(" %lf", &d);
  if (d > 0 && d < 4.0) {
    printf("N\n");
    return;
  }
  double low = 0, hi = d, mid;
  while (true) {
    mid = (low + hi) / 2;
    if (abs((mid * (d - mid)) - d) <= 1e-9) {
      printf("Y %.9lf %.9lf\n", mid, d - mid);
      return;
    }
    if ((mid * (d - mid)) > d) {
      hi = mid;
    } else
      low = mid;
  }
}
int main() {
  int tc;
  cin >> tc;
  for (int i = 0; i < tc; i++) {
    solve();
  }
}
