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;
long long int p[200008] = {0};
int main() {
  ios_base::sync_with_stdio(false);
  long long int n, z, x, i;
  long double p, q;
  cin >> n;
  for (i = 0; i < n; i++) {
    cin >> x;
    z = x * x - 4 * x;
    if (z < 0) {
      cout << "N\n";
      continue;
    }
    cout << "Y ";
    p = (long double)(x + sqrt(z)) / 2;
    q = (long double)(x - sqrt(z)) / 2;
    cout << setprecision(20) << p << " " << setprecision(20) << q << endl;
  }
}
