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;
const long long MaxN = 3e18;
const int MAXN = 120 * 1000;
int cD(long long n) { return floor(log10(n) + 1); }
long long last = 1000000000007;
long long p, sum = 0, sum1 = 0, sum2 = 0, i, t, h, j, g, k, ans = 0;
long long d1 = 0, d2 = 0;
long long x, q, l, y, l1, l2, r, m;
long long li, mi, hi;
string s1 = "312831303130313130313031";
string s;
long long c, c2;
int main() {
  ios_base::sync_with_stdio(false);
  cin >> t;
  while (t--) {
    double n, mid = 0.0;
    cin >> n;
    double l = 0.0, h = 2000.0;
    if (n == 0) {
      cout << "Y 0.0 0.0\n";
    } else if (n < 4) {
      cout << "N\n";
    } else if (n == 4) {
      cout << "Y 2.0 2.0\n";
    } else {
      for (int i = 0; i < 100; i++) {
        mid = l + (h - l) / 2;
        if (((mid * (n - mid)) <= n && (mid + (n - mid) <= n)))
          h = mid;
        else
          l = mid;
      }
      cout << "Y " << fixed << setprecision(10) << h << " " << n - h << "\n";
    }
  }
  return 0;
}
