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>
int main() {
  double d, a1, a2, b, c, det;
  int t;
  scanf("%d", &t);
  while (t--) {
    scanf("%lf", &d);
    b = -d;
    c = d;
    det = b * b - c * 4;
    if (det < 0) {
      printf("N\n");
      continue;
    }
    a1 = (-b + sqrt(det)) / 2;
    a2 = (-b - sqrt(det)) / 2;
    if (a1 >= 0 && a1 <= d) {
      printf("Y %.10lf %.10lf\n", a1, d - a1);
    } else if (a2 >= 0 && a2 <= d) {
      printf("Y %.10lf %.10lf\n", a2, d - a2);
    } else
      printf("N\n");
  }
  return 0;
}
