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;
int a[1000007], i, j, n;
string s;
int main() {
  int t;
  cin >> t;
  while (t--) {
    cin >> n;
    double lo = 0, hi = n;
    double mid;
    int ok = 0;
    for (int i = 0; i < 20000; i++) {
      mid = (lo + hi) / 2.0;
      double x = n - mid;
      if (fabs(n - mid * x) <= 0.00000001) {
        cout << "Y"
             << " " << setprecision(11) << x << " " << mid << endl;
        ok = 1;
        break;
      }
      if (x * mid > n)
        hi = mid;
      else
        lo = mid;
    }
    if (ok == 0) cout << "N" << endl;
  }
}
