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 double EPS = 1e-8;
int main() {
  int _T;
  scanf("%d", &_T);
  while (_T--) {
    int d;
    scanf("%d", &d);
    if (d == 0) {
      printf("Y %.9lf %.9lf\n", 0.0, 0.0);
      continue;
    }
    int delta = d * d - 4 * d;
    if (delta < 0) {
      printf("N\n");
      continue;
    }
    double r1 = (1.0 * d + sqrt(delta)) / 2.0;
    double x1 = d / r1;
    if (r1 < x1) swap(r1, x1);
    if (r1 >= EPS && x1 >= EPS) {
      printf("Y %.9lf %.9lf\n", r1, x1);
      continue;
    }
    double r2 = (d - sqrt(delta)) / 2;
    double x2 = d / r2;
    if (r2 >= EPS && x2 >= EPS) {
      printf("Y %.9lf %.9lf\n", r2, x2);
      continue;
    }
    printf("N\n");
  }
}