#include <bits/stdc++.h>
using namespace std;
double m = 0.000000001;
double bin_sear(double n, double s, double e, double prev) {
  if (s <= e) {
    double mid = (s + e) / 2;
    if (abs(mid - prev) <= m) return mid;
    prev = mid;
    double a = n - mid;
    if (a * mid == n) return mid;
    if (a * mid <= n) return bin_sear(n, s, mid, prev);
    return bin_sear(n, mid, e, prev);
  }
}
signed main() {
  long long t, i, j;
  cin >> t;
  double n;
  while (t--) {
    cin >> n;
    double ans = bin_sear(n, 0, n, -1);
    if (n <= 3 && n >= 1)
      cout << "N\n";
    else
      cout << "Y " << fixed << setprecision(10) << ans << " " << fixed
           << setprecision(10) << n - ans << "\n";
  }
  return 0;
}

### Prompt

Please create a solution in CPP 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>
using namespace std;
double m = 0.000000001;
double bin_sear(double n, double s, double e, double prev) {
  if (s <= e) {
    double mid = (s + e) / 2;
    if (abs(mid - prev) <= m) return mid;
    prev = mid;
    double a = n - mid;
    if (a * mid == n) return mid;
    if (a * mid <= n) return bin_sear(n, s, mid, prev);
    return bin_sear(n, mid, e, prev);
  }
}
signed main() {
  long long t, i, j;
  cin >> t;
  double n;
  while (t--) {
    cin >> n;
    double ans = bin_sear(n, 0, n, -1);
    if (n <= 3 && n >= 1)
      cout << "N\n";
    else
      cout << "Y " << fixed << setprecision(10) << ans << " " << fixed
           << setprecision(10) << n - ans << "\n";
  }
  return 0;
}

```