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;
#pragma GCC target("avx2")
#pragma GCC optimization("O3")
#pragma GCC optimization("unroll-loops")
bool sortbysec(const pair<int, int> &a, const pair<int, int> &b) {
  return (a.second < b.second);
}
const long long INF = 1e10 + 1;
const int mod = 1e9 + 7;
inline int add(int x, int y) {
  x += y;
  if (x >= mod) return x - mod;
  return x;
}
inline int sub(int x, int y) {
  x -= y;
  if (x < 0) return x + mod;
  return x;
}
inline int mul(int x, int y) { return (x * 1ll * y) % mod; }
inline int power(int x, int y) {
  int ans = 1;
  while (y) {
    if (y & 1) ans = mul(ans, x);
    x = mul(x, x);
    y >>= 1;
  }
  return ans;
}
inline int inv(int x) { return power(x, mod - 2); }
const long long FACTORIAL_SIZE = 1.1e6;
long long fact[FACTORIAL_SIZE], ifact[FACTORIAL_SIZE];
void gen_factorial(long long n) {
  fact[0] = fact[1] = ifact[0] = ifact[1] = 1;
  for (long long i = 2; i <= n; i++) {
    fact[i] = (i * fact[i - 1]) % mod;
  }
  ifact[n] = inv(fact[n]);
  for (long long i = n - 1; i >= 2; i--) {
    ifact[i] = ((i + 1) * ifact[i + 1]) % mod;
  }
}
long long nck(long long n, long long k) {
  long long den = (ifact[k] * ifact[n - k]) % mod;
  return (den * fact[n]) % mod;
}
const int N = 2e5 + 5;
long long d;
bool check(double a, double b) {
  if (abs(a + b - a * b) <= 0.000001 && abs((a + b) - d) <= 0.000001)
    return true;
  return false;
}
void solve() {
  cin >> d;
  cout << fixed << setprecision(10);
  if (d == 0) {
    cout << "Y"
         << " ";
    double x = 0;
    cout << x << " " << x << "\n";
    return;
  }
  for (long long i = 1; i * i <= d; i++) {
    if (d % i) continue;
    if (d * d - 4 * d < 0) continue;
    long long f = i;
    long long s = d / i;
    double x1 = (d + sqrt(d * d - 4 * d)) / (2.0 * s);
    double a1 = f / x1;
    double b1 = s * x1;
    if (check(a1, b1)) {
      cout << "Y " << a1 << " " << b1 << "\n";
      return;
    }
    swap(f, s);
    double x2 = (d + sqrt(d * d - 4 * d)) / (2.0 * s);
    double a2 = f / x2;
    double b2 = s * x2;
    if (check(a2, b2)) {
      cout << "Y " << a2 << " " << b2 << "\n";
      return;
    }
  }
  cout << "N"
       << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long tc = 1;
  cin >> tc;
  long long i = 1;
  while (tc--) {
    solve();
    i++;
  }
  return 0;
}
