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;
void FAST() { ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0); }
int main() {
  FAST();
  long long t;
  cin >> t;
  while (t-- != 0) {
    long double n;
    cin >> n;
    if (n * n - 4 * n < 0) {
      cout << "N" << endl;
      continue;
    }
    long double b = (n - sqrt(n * (n - 4))) / 2;
    long double a = n - b;
    if (a < 0 || b < 0) {
      cout << "N" << endl;
      continue;
    }
    cout << "Y " << fixed << setprecision(10) << a << " " << b << endl;
  }
  return 0;
}
struct Query {
  long long L, R, idx;
};
bool compare(long long a0, long long a1) { return (a0 > a1); }
long long power(long long x, long long y, long long p) {
  long long ans = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) ans = (ans * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return ans;
}
long long x, y, g;
void extendedEuclid(long long a, long long b) {
  if (b == 0) {
    g = a;
    x = 1;
    y = 0;
  } else {
    extendedEuclid(b, a % b);
    long long temp = x;
    x = y;
    y = temp - (a / b) * y;
  }
}
long long moduloInverse(long long a, long long p) {
  extendedEuclid(a, p);
  if (g == 1 && p != 1)
    return (x + p) % p;
  else
    return -1;
}
