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>
struct Roots {
  char YorN;
  double a;
  double b;
};
struct Roots GetNumbers(int d) {
  if (d * d - 4 * d < 0) {
    struct Roots roots = {'N', -1, -1};
    return roots;
  }
  double l = 0;
  double r = sqrt(d);
  double mid = (r + l) / 2;
  for (int i = 0; i < 500; ++i) {
    mid = (r + l) / 2;
    if (mid * (d - mid) > d) {
      r = mid;
    } else {
      l = mid;
    }
  }
  struct Roots roots = {'Y', mid, d - mid};
  return roots;
}
int main() {
  int t = 0;
  scanf("%d", &t);
  struct Roots* d_sequence = (struct Roots*)malloc(t * sizeof(struct Roots));
  for (int i, d = 0; i < t; ++i) {
    scanf("%d", &d);
    d_sequence[i] = GetNumbers(d);
  }
  for (int i = 0; i < t; ++i) {
    if (d_sequence[i].a == -1.0) {
      printf("%c\n", d_sequence[i].YorN);
    } else {
      printf("%c %.9f %.9f\n", d_sequence[i].YorN, d_sequence[i].a,
             d_sequence[i].b);
    }
  }
  return 0;
}
