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;
double n;
void bsearch(double low, double high) {
  double mid, ans = 0;
  while (high - low > .0000000001) {
    mid = (low + high) / 2.0;
    double x = (mid * n) - (mid * mid);
    if (x >= n) {
      ans = mid;
      high = mid;
    } else {
      low = mid;
    }
  }
  double ck = ans * (n - ans);
  if ((n - ck) > .000001) {
    printf("N\n");
  } else {
    printf("Y %.10f ", abs(n - ans));
    printf("%.10f\n", ans);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    cin >> n;
    if (n == 0) {
      double a = 0;
      printf("Y %.10f ", a);
      printf("%.10f\n", a);
    } else {
      bsearch(0, n);
    }
  }
  return 0;
}
