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() {
  long long int t;
  cin >> t;
  if (t == 0) {
    cout << "Y 0.00000000 0.00000000" << endl;
    return;
  }
  if (t < 4) {
    cout << "N" << endl;
    return;
  }
  long double d = (long double)t;
  long double l = 0, r = d, m = (l + r) / 2;
  while (abs(d - ((m * m) / (m - 1))) > 0.00000000001) {
    m = (l + r) / 2;
    if (d - ((m * m) / (m - 1)) < 0) {
      r = m;
    } else {
      l = m;
    }
  }
  printf("Y %.10Lf %.10Lf\n", m, d - m);
}
int main() {
  int t;
  cin >> t;
  while (t--) solve();
  return 0;
}
