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 int maxn = 2e5 + 100;
const int inf = 1e9 + 7;
const long long mod = 1e9 + 7;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  for (int tc = 0; tc < t; tc++) {
    int d;
    cin >> d;
    if (d == 1 || d == 2 || d == 3) {
      cout << "N" << endl;
      continue;
    }
    double lo = 0;
    double hi = d / 2;
    for (int i = 0; i < 100; i++) {
      double mid = (lo + hi) / 2;
      if (mid * (d - mid) < d) {
        lo = mid;
      } else {
        hi = mid;
      }
    }
    cout << fixed << setprecision(12) << "Y " << d - lo << " " << lo << endl;
  }
}
