#include <bits/stdc++.h>
using namespace std;
int d, T;
double delta;
int main() {
  scanf("%d", &T);
  while (T--) {
    scanf("%d", &d);
    if (d < 4 && d != 0) {
      printf("N\n");
      continue;
    }
    printf("Y ");
    delta = 1.0 * sqrt(d * d - d * 4);
    printf("%.9lf %.9lf\n", (d - delta) * 1.0 / 2, (delta + d) * 1.0 / 2);
  }
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

Try guessing the statement from this picture: 

<image>

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int d, T;
double delta;
int main() {
  scanf("%d", &T);
  while (T--) {
    scanf("%d", &d);
    if (d < 4 && d != 0) {
      printf("N\n");
      continue;
    }
    printf("Y ");
    delta = 1.0 * sqrt(d * d - d * 4);
    printf("%.9lf %.9lf\n", (d - delta) * 1.0 / 2, (delta + d) * 1.0 / 2);
  }
  return 0;
}

```