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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
string s, s1;
double n, m, a, b, i, j, d, t, t1, cs = 0, counT = 0, k, ans = 0, l = 0,
                                   temp = 0, sum = 0, Max, Min, num;
long long int arr[10000000];
vector<long long int> vc;
map<long long int, long long int> mp;
int main() {
  cin >> n;
  while (n--) {
    cin >> d;
    temp = d * d - 4 * d;
    if (temp < 0) {
      cout << "N" << '\n';
      continue;
    }
    t = .5 * (d + sqrt((d * d - ((4 * d)))));
    t1 = .5 * (d - sqrt((d * d - ((4 * d)))));
    Min = min(t, t1);
    cout << "Y"
         << " " << fixed << setprecision(9) << d - Min << " " << fixed
         << setprecision(9) << Min << '\n';
  }
}