Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int main() {
  int i, j, t1, t2, t3, t4, t;
  double buf1, buf2, buf3, buf4, a1, a2, d;
  scanf("%d", &t);
  while (t--) {
    scanf("%lf", &d);
    buf1 = d * d - (4.0 * d);
    if (buf1 < 0) {
      printf("N\n");
      continue;
    }
    buf1 = sqrt(buf1);
    a1 = (d + buf1) / 2.0;
    a2 = (d - buf1) / 2.0;
    if (a1 < 0 || a2 < 0) {
      printf("N\n");
      continue;
    }
    printf("Y ");
    printf("%.15lf %.15lf\n", a1, a2);
  }
  return 0;
}