#include <bits/stdc++.h>
using namespace std;
void solve() {
  int d;
  cin >> d;
  double L = 0, R = d, mid;
  double big = 0, small = 0;
  if (d == 4) {
    printf("Y %.9lf %.9lf\n", 2.000000000, 2.000000000);
    return;
  } else if (d == 0) {
    printf("Y %.9lf %.9lf\n", 0.000000000, 0.000000000);
    return;
  } else if (d < 4) {
    printf("N\n");
    return;
  }
  while (fabs(d - (big * small)) >= 1e-7) {
    mid = L + (R - L) / 2;
    big = mid;
    small = d - mid;
    if (big * small > d)
      L = mid;
    else
      R = mid;
  }
  printf("Y %.9lf %.9lf\n", big, small);
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    solve();
  };
}

### Prompt

Create a solution in cpp for 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>
using namespace std;
void solve() {
  int d;
  cin >> d;
  double L = 0, R = d, mid;
  double big = 0, small = 0;
  if (d == 4) {
    printf("Y %.9lf %.9lf\n", 2.000000000, 2.000000000);
    return;
  } else if (d == 0) {
    printf("Y %.9lf %.9lf\n", 0.000000000, 0.000000000);
    return;
  } else if (d < 4) {
    printf("N\n");
    return;
  }
  while (fabs(d - (big * small)) >= 1e-7) {
    mid = L + (R - L) / 2;
    big = mid;
    small = d - mid;
    if (big * small > d)
      L = mid;
    else
      R = mid;
  }
  printf("Y %.9lf %.9lf\n", big, small);
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    solve();
  };
}

```