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 long long MAX = 1LL * 1e10 + 1;
const long long MOD = 1LL * 1e9 + 7;
const string alpf = "abcdefghijklmnopqrstuvwxyz";
inline void files() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
long long n, m, ans;
int main() {
  files();
  cin >> n;
  cout << setprecision(9);
  cout << fixed;
  for (int i = 0; i < n; ++i) {
    long double test;
    cin >> test;
    if (!test) {
      cout << "Y " << 0 << ' ' << 0 << "\n";
      continue;
    }
    if (test < 4) {
      cout << "N\n";
      continue;
    }
    bool check = 0;
    long double l = 1, r = 2;
    int cur = 0;
    long double res;
    while (cur < 50) {
      long double mid = (l + r) / 2;
      if (mid * (test - mid) <= test)
        l = mid;
      else
        r = mid;
      cur++;
      res = mid;
    }
    cout << "Y " << res << ' ' << test - res << '\n';
  }
}