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;
void k_007() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
bool sortbysec(const pair<long long, long long> &a,
               const pair<long long, long long> &b) {
  return (a.second < b.second);
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long prime(long long n) {
  long long flag = 1;
  for (long long i = 2; i <= sqrt(n); i++) {
    if (n % i == 0) {
      flag = 0;
      break;
    }
  }
  return flag;
}
long long power(long long x, long long y) {
  if (y == 0)
    return 1;
  else if (y % 2 == 0)
    return ((power(x, y / 2) % 1000000007) * (power(x, y / 2) % 1000000007)) %
           1000000007;
  else
    return ((x % 1000000007) * (power(x, y / 2) % 1000000007) *
            (power(x, y / 2) % 1000000007)) %
           1000000007;
}
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
int32_t main() {
  k_007();
  long long x;
  cin >> x;
  while (x--) {
    long double d;
    cin >> d;
    long double val = d * d - 4 * d;
    if (val < 0) {
      cout << "N" << endl;
      continue;
    }
    long double a = (d + sqrt(val)) / (long double)2;
    long double b = 0;
    if (d == 0) {
      cout << "Y " << fixed << setprecision(9) << a << " " << fixed
           << setprecision(9) << b << endl;
      continue;
    }
    b = d / (long double)a;
    cout << "Y " << fixed << setprecision(9) << a << " " << fixed
         << setprecision(9) << b << endl;
  }
}
