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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 N = 1e6 + 5;
long long T, d;
string s;
int32_t main() {
  cin >> T;
  while (T--) {
    cin >> d;
    if (d > 0 && d <= 3) {
      puts("N");
      continue;
    }
    printf("Y");
    long double a1 = d / 2.0 - sqrt((long double)d * d / 4.0 - d), b1 = d - a1;
    long double a2 = d / 2.0 + sqrt((long double)d * d / 4.0 - d), b2 = d - a1;
    if (a1 >= 0 && a2 >= 0)
      printf(" %.20Lf %.20Lf\n", a1, b1);
    else
      printf(" %.20Lf %.20Lf\n", a2, b2);
  }
}
