Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5, M = N, inf = (int)1e9;
const int mod = 1e9 + 7;
int fastAbs(int n) { return (n ^ (n >> 31)) - (n >> 31); }
long long multiply(long long a, long long b) {
  return ((a % mod) * (b % mod)) % mod;
}
long long add(long long a, long long b) {
  return ((a % mod) + (b % mod)) % mod;
}
long long sub(long long a, long long b) {
  return ((a % mod) - (b % mod) + mod) % mod;
}
long long ans;
map<pair<int, int>, int> mp;
const int dx[4] = {1, 0, 0, -1};
const int dy[4] = {0, -1, 1, 0};
void solve() {
  long double d;
  cin >> d;
  long double a1, b1, a2, b2;
  long double c = (1.0 * d * d) - (4.0 * d);
  cerr << c << " ";
  if (c >= 0) {
    a1 = (d + sqrt(c)) / 2.0;
    a2 = (d - sqrt(c)) / 2.0;
    b1 = d - a1;
    b2 = d - a2;
    if (a1 >= 0 && b1 >= 0) {
      cout << "Y " << fixed << setprecision(9) << a1 << " " << b1 << "\n";
      return;
    }
    if (a2 >= 0 && b2 >= 0) {
      cout << "Y " << fixed << setprecision(9) << a2 << " " << b2 << "\n";
      return;
    } else
      cout << "N\n";
  } else
    cout << "N\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int q = 1;
  cin >> q;
  while (q--) solve();
}