#include <bits/stdc++.h>
const int MAXN = 1e5 + 10;
using namespace std;
map<int, int>::iterator it;
int str[MAXN];
int main() {
  int T;
  double m, n;
  cin >> T;
  while (T--) {
    cin >> m;
    int flag = 0;
    if (int(m * m - 4 * m) < 0)
      printf("N\n");
    else {
      double a, b;
      a = (m + sqrt(m * m - 4 * m)) / 2;
      b = (m - sqrt(m * m - 4 * m)) / 2;
      printf("Y %.9lf %.9lf\n", a, b);
    }
  }
}

### Prompt

Your challenge is to write a CPP solution to the following problem:

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>
const int MAXN = 1e5 + 10;
using namespace std;
map<int, int>::iterator it;
int str[MAXN];
int main() {
  int T;
  double m, n;
  cin >> T;
  while (T--) {
    cin >> m;
    int flag = 0;
    if (int(m * m - 4 * m) < 0)
      printf("N\n");
    else {
      double a, b;
      a = (m + sqrt(m * m - 4 * m)) / 2;
      b = (m - sqrt(m * m - 4 * m)) / 2;
      printf("Y %.9lf %.9lf\n", a, b);
    }
  }
}

```