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 double err = 0.000000001;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int tc;
  cin >> tc;
  while (tc--) {
    long double n;
    cin >> n;
    long double low = 0, high = n, mid, a = 0, b = 0;
    if (n == 0)
      cout << "Y " << 0.0 << " " << 0.0 << endl;
    else if (n < 4)
      cout << "N" << endl;
    else {
      while (!(abs((a * b) - n) < err && abs((a + b) - n) < err)) {
        mid = (low + high) / 2;
        a = n - mid;
        b = mid;
        if (a * b > n)
          low = mid;
        else
          high = mid;
      }
      cout << fixed << setprecision(10) << "Y " << b << " " << a << endl;
    }
  }
  return 0;
}
