Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int inf = 1e9 + 7;
const int mod = 998244353;
const int maxn = 1e5 + 1;
long long int sqr(long long int x) { return x * x; }
long long int modexpo(long long int a, long long int b) {
  a %= inf;
  long long int res = 1;
  while (b > 0) {
    if (b & 1) res = res * a % inf;
    a = a * a % inf;
    b >>= 1;
  }
  return res;
}
void solve() {
  long long int n;
  cin >> n;
  if (n == 0) {
    cout << "Y 0.000000000 0.000000000" << endl;
  } else if (n < 4 && n != 0) {
    cout << "N" << endl;
  } else {
    long long int z = n * n - 4 * n;
    double k = double(sqrt(z));
    double k1 = (double(n) + k) / (1.0 * 2);
    double k2 = (double(n) - k) / (1.0 * 2);
    cout << "Y"
         << " " << fixed << setprecision(12) << k1 << " " << fixed
         << setprecision(12) << k2 << endl;
  }
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
}