#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
const int maxn = 1e3 + 10;
const int inf = 0x7fffffff;
const long long INF = 9E18;
const int mod = 1e9 + 7;
const long long mod2 = 1e9 + 9;
const double eps = 1e-7;
const double pi = acos(-1.0);
template <typename T>
inline bool read(T &x) {
  T f = 1;
  char ch = getchar();
  if (ch == EOF) return false;
  for (; ch < '0' || ch > '9'; ch = getchar())
    if (ch == '-') f = -1;
  for (x = 0; ch >= '0' && ch <= '9'; ch = getchar()) x = x * 10 + ch - '0';
  x *= f;
  return true;
}
int dcmp(double x) { return fabs(x) <= eps ? 0 : (x > 0 ? 1 : -1); }
int main() {
  int test;
  cin >> test;
  while (test--) {
    double x;
    cin >> x;
    if (dcmp(x) == 0)
      cout << "Y 0 0" << endl;
    else {
      if (x * x - 4 * x < 0)
        cout << "N" << endl;
      else {
        double res = (x + sqrt(x * x - 4 * x)) / 2;
        double a = x / res;
        cout << fixed << setprecision(10) << "Y " << a << " " << res << endl;
      }
    }
  }
  return 0;
}

### Prompt

In cpp, your task is to solve 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>
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
const int maxn = 1e3 + 10;
const int inf = 0x7fffffff;
const long long INF = 9E18;
const int mod = 1e9 + 7;
const long long mod2 = 1e9 + 9;
const double eps = 1e-7;
const double pi = acos(-1.0);
template <typename T>
inline bool read(T &x) {
  T f = 1;
  char ch = getchar();
  if (ch == EOF) return false;
  for (; ch < '0' || ch > '9'; ch = getchar())
    if (ch == '-') f = -1;
  for (x = 0; ch >= '0' && ch <= '9'; ch = getchar()) x = x * 10 + ch - '0';
  x *= f;
  return true;
}
int dcmp(double x) { return fabs(x) <= eps ? 0 : (x > 0 ? 1 : -1); }
int main() {
  int test;
  cin >> test;
  while (test--) {
    double x;
    cin >> x;
    if (dcmp(x) == 0)
      cout << "Y 0 0" << endl;
    else {
      if (x * x - 4 * x < 0)
        cout << "N" << endl;
      else {
        double res = (x + sqrt(x * x - 4 * x)) / 2;
        double a = x / res;
        cout << fixed << setprecision(10) << "Y " << a << " " << res << endl;
      }
    }
  }
  return 0;
}

```