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 inf = 0x3f3f3f3f;
const double e = 1e-6;
int main() {
  int T;
  double n;
  scanf("%d", &T);
  while (T--) {
    scanf("%lf", &n);
    double a = 0, b = n;
    bool flag = false;
    for (int i = 0; i < 10000; i++) {
      double m = (a + b) / 2, d = n - m;
      if (m * d - m - d > 1e-9)
        b = m;
      else if (m + d - d * m > 1e-9)
        a = m;
    }
    double m = (a + b) / 2, d = n - m;
    if (m + d - n <= e && m + d - n >= -e && (m + d) - m * d <= e &&
        (m + d) - m * d >= -e && m * d - n <= e && m * d - n >= -e)
      printf("Y %.9lf %.9lf\n", d, m);
    else
      puts("N");
  }
  return 0;
}
