Problem: Try guessing the statement from this picture: 

<image>

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 long long N = 1e6 + 5;
long long max(long long a, long long b) {
  if (a >= b)
    return a;
  else
    return b;
}
long long min(long long a, long long b) {
  if (a >= b)
    return b;
  else
    return a;
}
long long diff(long long a, long long b) {
  if (a >= b)
    return a - b;
  else
    return b - a;
}
long long mod(long long a) {
  if (a >= 0)
    return a;
  else
    return -a;
}
void pairsort(long long a[], long long b[], long long n) {
  pair<long long, long long> pairt[n];
  for (long long i = 0; i < n; i++) {
    pairt[i].first = a[i];
    pairt[i].second = b[i];
  }
  sort(pairt, pairt + n);
  for (long long i = 0; i < n; i++) {
    a[i] = pairt[i].first;
    b[i] = pairt[i].second;
  }
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long rev(long long n) {
  long long ans = 0;
  while (n) {
    ans = ans * 10 + n % 10;
    n = n / 10;
  }
  return ans;
}
long long fact(long long n) {
  if (n == 0) return 1;
  if (n == 1) return 1;
  if (n >= 2) return n * fact(n - 1);
}
long long isPrime(long long n) {
  for (long long i = 2; i * i <= n; i++)
    if (n % i == 0) return 0;
  return 1;
}
long long solve(long long n, long long m) {}
long long C(long long n, long long r) {
  if (r > n - r) r = n - r;
  long long ans = 1;
  long long i;
  for (i = 1; i <= r; i++) {
    ans *= n - r + i;
    ans /= i;
  }
  return ans;
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long t;
  cin >> t;
  while (t--) {
    double d;
    cin >> d;
    if (d * d - 4 * d >= 0) {
      cout << "Y"
           << " ";
      double x = d + sqrt(d * d - 4 * d);
      x = x / 2;
      cout << setprecision(10) << x << " " << d - x << "\n";
    } else
      cout << "N"
           << "\n";
  }
  return 0;
}