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;
template <class T>
T gcd(T a, T b) {
  return (b != 0 ? gcd<T>(b, a % b) : a);
}
template <class T>
T lcm(T a, T b) {
  return (a / gcd<T>(a, b) * b);
}
double tick() {
  static clock_t oldt, newt = clock();
  double diff = 1.0 * (newt - oldt) / CLOCKS_PER_SEC;
  oldt = newt;
  return diff;
}
float roundoff(float val, float p) {
  return floor(val * pow(10.0, p) + 0.5) / pow(10.0, p);
}
long long int power(long long int a, long long int b, long long int mod) {
  long long int x = 1, y = a;
  while (b > 0) {
    if (b & 1) {
      x = (x * y) % mod;
    }
    y = (y * y) % mod;
    b /= 2;
  }
  return x;
}
const int INF = 0x3f3f3f3f;
const float pi = 3.14159265;
int main(void) {
  ios_base::sync_with_stdio(true);
  cin.tie(NULL);
  cout.tie(NULL);
  int t;
  scanf("%d", &t);
  while (t--) {
    double nn;
    int n;
    cin >> n;
    nn = n * 1.0;
    if (n > 0 and n < 4) {
      cout << 'N' << '\n';
    } else {
      double a = (nn - sqrt(nn * nn - 4 * nn)) / 2.0,
             b = (nn + sqrt(nn * nn - 4 * nn)) / 2.0;
      cout << "Y ";
      cout << fixed << setprecision(9) << a << ' ' << b << '\n';
    }
  }
  return 0;
}
