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;
mt19937_64 rang(
    chrono::high_resolution_clock::now().time_since_epoch().count());
int rng(int lim) {
  uniform_int_distribution<int> uid(0, lim - 1);
  return uid(rang);
}
int i, j;
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  srand(chrono::high_resolution_clock::now().time_since_epoch().count());
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    double l = 0;
    double r = n;
    for (i = 0; i < 100; i++) {
      double m = (l + r) / 2;
      double mm = n - m;
      if (m * mm < n) {
        r = m;
      } else {
        l = m;
      }
    }
    if (n == 0) {
      cout << "Y"
           << " " << 0.0000000000 << " " << 0.0000000000 << "\n";
    } else if (l == 0) {
      cout << "N"
           << "\n";
    } else {
      cout << "Y"
           << " " << setprecision(20) << l << " " << setprecision(20) << n - l
           << "\n";
    }
  }
}
