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;
const long long int mod = 1000000007;
using namespace std;
void solve() {
  long long int n, t, i, x, y, z, k, mx = 0, mn = 99999999, count = 0, pos = 0,
                                     flag = 0, sm = 0, a, b;
  cin >> t;
  while (t) {
    long double d;
    cin >> d;
    long double ans;
    ans = (d + sqrt(d * d - 4 * d)) / 2;
    long double num1 = ans;
    long double num2 = d - ans;
    if (d * d - 4 * d < 0)
      cout << "N" << endl;
    else {
      cout << "Y"
           << " " << fixed << setprecision(9) << num1 << " " << fixed
           << setprecision(9) << num2 << endl;
    }
    t--;
  }
}
int main() { solve(); }
