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;
const int INF = 1000000000;
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    int d;
    scanf("%d", &d);
    double a, b;
    if (d == 4) {
      a = 2;
      b = 2;
      printf("Y %.9lf %.9lf\n", b, a);
      continue;
    }
    double lo = 0, hi = 1000;
    for (int i = 0; i < 100; i++) {
      b = (lo + hi) / 2;
      if (b * b - (d * b) + (double)d > 0)
        hi = b;
      else
        lo = b;
    }
    a = (double)d / b;
    int tmp1 = a + b;
    int tmp2 = a * b;
    int tmp3 = a * b - 0.000001;
    int tmp4 = a * b + 0.000001;
    if (tmp1 == tmp2 || tmp1 == tmp3 || tmp1 == tmp4)
      printf("Y %.9lf %.9lf\n", b, a);
    else
      puts("N");
  }
  return 0;
}
