#include <bits/stdc++.h>
using namespace std;
vector<long long> v;
pair<long long, long long> p;
map<long long, long long> mp;
set<long long> s;
long long mn = INT_MAX, mx = INT_MIN;
long long n, m, tc, i, j, tmp, sum, cn, ans, res, pos, flag;
long long a[200007];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> tc;
  while (tc--) {
    double d;
    cin >> d;
    double det = d * d - 4 * d;
    if (det < 0)
      puts("N");
    else {
      double x = (d + sqrt(det)) / 2.0;
      double y = (d - sqrt(det)) / 2.0;
      printf("Y %.9f %.9f\n", x, y);
    }
  }
  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;
vector<long long> v;
pair<long long, long long> p;
map<long long, long long> mp;
set<long long> s;
long long mn = INT_MAX, mx = INT_MIN;
long long n, m, tc, i, j, tmp, sum, cn, ans, res, pos, flag;
long long a[200007];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> tc;
  while (tc--) {
    double d;
    cin >> d;
    double det = d * d - 4 * d;
    if (det < 0)
      puts("N");
    else {
      double x = (d + sqrt(det)) / 2.0;
      double y = (d - sqrt(det)) / 2.0;
      printf("Y %.9f %.9f\n", x, y);
    }
  }
  return 0;
}

```