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;
const double eps = 1e-7;
void solve(int x) {
  double l, r, a, b;
  int flag = 0;
  l = 0.0;
  r = 1.0 * x / 2.0;
  while (l < r) {
    double mid = (l + r) / 2.0;
    a = mid;
    b = 1.0 * x - mid;
    if (fabs((a + b) - a * b) <= eps) {
      flag = 1;
      cout << "Y"
           << " ";
      printf("%.9f %.9f\n", b, a);
      break;
    }
    if ((a + b) <= a * b)
      r = mid;
    else
      l = mid;
  }
  if (flag == 0) cout << "N" << endl;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    int d;
    cin >> d;
    if (d == 0) {
      double b = 0.0, a = 0.0;
      cout << "Y"
           << " ";
      printf("%.9f %.9f\n", b, a);
    } else
      solve(d);
  }
  return 0;
}
