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 n, i, x, de;
double a, b1, b2;
int main() {
  cin >> n;
  for (i = 1; i <= n; i++) {
    cin >> x;
    de = x * x - 4 * x;
    if (de < 0)
      cout << "N\n";
    else {
      b1 = (x + sqrt(de)) / 2;
      b2 = (x + sqrt(de)) / 2;
      if (b1 >= 0) {
        a = x - b1;
        if (a >= 0) cout << setprecision(12) << "Y " << a << " " << b1 << "\n";
      } else {
        if (b2 >= 0) a = x - b2;
        if (a >= 0) cout << setprecision(12) << "Y " << a << " " << b2 << "\n";
      }
    }
  }
  return 0;
}
