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;
long long const mod = 1e9 + 7;
long long power(long long x, long long y, long long mod = 2e18) {
  long long ans = 1;
  x %= mod;
  while (y) {
    if (y & 1) ans = (x * ans) % mod;
    x = (x * x) % mod;
    y >>= 1;
  }
  return ans;
}
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    double high = n;
    double low = 0.0;
    double mid;
    for (long long i = 0; i < 300; i++) {
      mid = (high + low) / 2.0;
      if (mid * mid / (mid - 1) > n)
        high = mid;
      else
        low = mid;
    }
    cout << setprecision(10);
    double b = mid / (mid - 1);
    if (b > n || mid > n) {
      cout << "N\n";
      continue;
    }
    cout << "Y " << mid << " " << b << endl;
  }
  return 0;
}
