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;
float power(float f, long long n) {
  float temp = power(f, n / 2);
  if (n % 2 == 0) {
    return temp * temp;
  }
  return temp * temp * f;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long tc;
  cin >> tc;
  while (tc--) {
    long long n;
    cin >> n;
    long long a = n * n - 4 * n;
    if (a < 0) {
      cout << "N"
           << "\n";
      continue;
    }
    long double d = pow(a, 0.5);
    long double d2 = (n - d) / 2;
    long double d1 = (n + d) / 2;
    cout << fixed << setprecision(11) << "Y " << d1 << " " << d2 << "\n";
  }
}
