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>
const double PI = acos(-1.0);
using namespace std;
void fast_in_out() {
  std::ios_base::sync_with_stdio(NULL);
  cin.tie(NULL);
  cout.tie(NULL);
}
void pause() {}
const int mod = 1e9 + 7;
int dx[] = {0, 0, -1, -1, -1, 1, 1, 1};
int dy[] = {-1, 1, 0, 1, -1, -1, 0, 1};
long long fast_power(long long base, long long power) {
  long long result = 1;
  while (power > 0) {
    if (power & 1) {
      result = (result * base) % 1000000007;
    }
    base = (base * base) % 1000000007;
    power >>= 1;
  }
  return result;
}
double eps = 1e-6;
int main() {
  fast_in_out();
  cout << fixed << setprecision(9);
  int t;
  cin >> t;
  while (t--) {
    long long d;
    cin >> d;
    if (d > 0 && d < 4)
      cout << "N\n";
    else {
      double a = (d + sqrt(d * (d - 4))) / 2.0;
      double b = (d - sqrt(d * (d - 4))) / 2.0;
      cout << "Y " << a << " " << b << '\n';
    }
  }
  pause();
  return 0;
}
