Try guessing the statement from this picture: 

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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;
void swap(long long int *x, long long int *y) {
  long long int z = *x;
  *x = *y;
  *y = z;
}
bool isprime(long long int n) {
  long long int i;
  for (i = 2; i * i <= n; i++) {
    if (n % i == 0) return 0;
  }
  return 1;
}
bool b(long long int n) {
  if (n != 1) {
    if (sqrt(n) == (int)sqrt(n)) {
      if (isprime(sqrt(n)))
        return 1;
      else
        return 0;
    } else
      return 0;
  } else
    return 0;
}
long long int index(long long int l, long long int a[], long long int x,
                    long long int n) {
  long long int i;
  for (i = l; i < n; i++) {
    if (a[i] > x) return i;
  }
  return n;
}
bool sortbysec(const pair<int, int> &a, const pair<int, int> &b) {
  return (a.second < b.second);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long int t;
  cin >> t;
  while (t--) {
    long long int d;
    double a, b;
    cin >> d;
    if ((d == 0) || (d >= 4)) {
      cout << "Y ";
      a = (d + sqrt(pow(d, 2) - 4 * d)) / 2;
      b = (d - sqrt(pow(d, 2) - 4 * d)) / 2;
      cout << fixed << setprecision(9) << a << " " << b;
    } else
      cout << "N";
    cout << endl;
  }
  return 0;
}
