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 Mod(long long x, long long y, long long p) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
class Triplet {
 public:
  long long x;
  long long y;
  long long gcd;
};
Triplet extendedEuclid(long long a, long long b) {
  if (b == 0) {
    Triplet ans;
    ans.gcd = a;
    ans.x = 1;
    ans.y = 0;
    return ans;
  }
  Triplet smallAns = extendedEuclid(b, a % b);
  Triplet ans;
  ans.gcd = smallAns.gcd;
  ans.x = smallAns.y;
  ans.y = smallAns.x - (a / b) * smallAns.y;
  return ans;
}
long long mmInverse(long long a, long long m) {
  Triplet ans = extendedEuclid(a, m);
  return ans.x;
}
set<long long> st;
void factorize(long long n) {
  long long count = 0;
  while (!(n % 2)) {
    n >>= 1;
    count++;
  }
  if (count) st.insert(2);
  for (long long i = 3; i <= sqrt(n); i += 2) {
    count = 0;
    while (n % i == 0) {
      count++;
      n = n / i;
    }
    if (count) st.insert(i);
  }
  if (n > 2) st.insert(n);
}
bool isPrime(long long n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t;
  cin >> t;
  while (t--) {
    long long d;
    cin >> d;
    long double x = d * d - 4 * d;
    if (x < 0) {
      cout << "N" << endl;
      continue;
    }
    long double rootx = sqrtl(x);
    if (d - rootx < 0) {
      cout << "N" << endl;
      continue;
    }
    cout << "Y"
         << " ";
    long double fir = (d - (rootx)) / 2;
    long double sec = (d + rootx) / 2;
    cout << fixed << setprecision(10) << fir << " ";
    cout << fixed << setprecision(10) << sec << endl;
  }
}
