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;
long long modx(long long Base, long long exponent) {
  long long ans = 1;
  if (Base == 1) return Base;
  while (exponent) {
    if (exponent & 1) ans = (ans * Base) % 1000000007LL;
    Base = (Base * Base) % 1000000007LL;
    exponent = exponent >> 1;
  }
  return ans;
}
long long inmodx(long long num) { return (modx(num, 1000000007LL - 2LL)); }
bool cmp(pair<pair<int, int>, int> a, pair<pair<int, int>, int> b) {
  if (a.first.second < b.first.second) return 1;
  if (a.first.second > b.first.second) return 0;
  if (a.first.first <= b.first.first) return 1;
  return 0;
}
const int N = (1e6) + 9;
const int M = (N << 2) + 9;
const int LOGN = ((int)log2(N)) + 3;
const int LOGM = ((int)log2(M)) + 3;
const int BUCK = 450;
void solve() {
  long long d, D;
  double a, b;
  cin >> d;
  D = d * d - 4LL * d;
  if (D < 0) {
    cout << "N\n";
    return;
  }
  a = d + sqrt(1.0 * D);
  b = d - sqrt(1.0 * D);
  a *= 0.5;
  b *= 0.5;
  cout << "Y " << a << " " << b << '\n';
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  clock_t clk;
  clk = clock();
  int t = 1, cs;
  cout << fixed << setprecision(9);
  cin >> t;
  for (cs = (1); cs < (t + 1); cs++) {
    solve();
  }
  clk = clock() - clk;
  cerr << fixed << setprecision(6) << "Time: " << ((double)clk) / CLOCKS_PER_SEC
       << "\n";
  return 0;
}