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 long long int MUL = 1000000000;
int main() {
  long long int t, z, cnt = 0;
  double lo = 0, mid, d, hi, a, b;
  cin >> t;
  for (int i = 0; i < t; i++) {
    cin >> z;
    if (z == 1) {
      cout << 'N' << endl;
      continue;
    }
    if (!z) {
      cout << "Y 0.0 0.0" << endl;
      continue;
    }
    d = (double)z;
    hi = d - 1.0;
    lo = d - 2.0;
    cnt = 0;
    while (1) {
      mid = (lo + hi) / 2.0;
      cnt++;
      if (cnt == 1000) break;
      a = mid;
      b = d - mid;
      if (a * b > d)
        lo = mid;
      else
        hi = mid;
      if (abs(d - (a * b)) < 0.000001 && abs(a * b - a - b) < 0.000001) break;
    }
    if (cnt == 1000)
      cout << "N\n";
    else
      cout << setprecision(13) << "Y " << a << ' ' << b << endl;
  }
  return 0;
}
