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;
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  if (b == 0) return a;
  return gcd(b % a, a);
}
long long power(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b % 2 != 0) res = (res * 1ll * a);
    b /= 2;
    a = (a * 1ll * a);
  }
  return res;
}
void solve() {
  int n;
  cin >> n;
  double ans = sqrt(n * n - 4 * n);
  double a = (n - ans) / double(2);
  double b = (n + ans) / double(2);
  if (n == 1 || n == 2 || n == 3) {
    cout << "N" << endl;
    return;
  }
  if (n == 0) {
    a = 0;
    b = 0;
  }
  cout << "Y " << fixed << setprecision(10) << a << " ";
  cout << fixed << setprecision(10) << b << endl;
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t;
  cin >> t;
  while (t--) solve();
  return 0;
}
