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 d;
double calc(double mid) { return 1 / mid + 1 / (double(d) - mid); }
int main() {
  std::ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    cin >> d;
    if (d == 0) {
      cout << "Y"
           << " " << 0 << " " << 0 << "\n";
      continue;
    }
    if (d < 4) {
      cout << "N"
           << "\n";
      continue;
    }
    cout << "Y"
         << " ";
    double dx = 0.0000000001;
    double l = 1 + dx, r = d / 2, mid;
    while (l <= r) {
      mid = (l + r) / 2.0;
      if (calc(mid) > 1)
        l = mid + dx;
      else
        r = mid - dx;
    }
    cout << fixed << setprecision(12) << l - dx << " " << d - l + dx << "\n";
  }
}
