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
#include <bits/stdc++.h>
using namespace std;
long long int gcd(long long int a, long long int b) {
  if (b != 0)
    return gcd(b, a % b);
  else
    return a;
}
long long int power(long long int a, long long int b) {
  if (b == 0) return 1;
  if (b == 1) return a;
  if (b % 2 == 0) {
    long long int t = power(a, b / 2);
    return t * t;
  } else {
    long long int t = power(a, b / 2);
    return a * t * t;
  }
}
long long int powin(long long int a, long long int b, long long int m) {
  if (b == 0)
    return 1;
  else if (b == 1)
    return a;
  else if (b % 2 == 0) {
    long long int t = powin(a, b / 2, m);
    return (t * t) % m;
  } else {
    long long int t = powin(a, b / 2, m);
    return (((t * t) % m) * a) % m;
  }
}
bool p[100009];
void fun() {
  for (long long int i = 2; i < 100009; i++) {
    if (p[i]) continue;
    for (long long int j = i * i; j < 100009; j += i) {
      p[j] = true;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  long long int tt;
  cin >> tt;
  while (tt--) {
    double d;
    cin >> d;
    if (d * d - 4 * d < 0) {
      cout << "N\n";
    } else {
      cout << "Y ";
      double b = -1 * d;
      double c = d;
      double a = 1;
      d = d * d - 4 * d;
      d = sqrt(d);
      double x = (-b + d) / (2 * a);
      double y = (-b - d) / (2 * a);
      cout << fixed << setprecision(10) << x;
      cout << " ";
      cout << fixed << setprecision(10) << y;
      cout << endl;
    }
  }
  return 0;
}
