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;
void solve(double d) {
  double a, b, mid, first = 0, last = d;
  int cnt = 1000;
  while (first <= last and cnt--) {
    mid = (first + last) / 2.0;
    double val = d - mid;
    a = mid + val;
    b = mid * val;
    if (abs(a - b) <= 0.00000001) {
      printf("Y %.9lf %.9lf\n", val, mid);
      return;
    }
    if (a > d || b > d)
      last = mid;
    else
      first = mid;
  }
  puts("N");
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    double d;
    cin >> d;
    solve(d);
  }
  return 0;
}