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;
const long long mos = 0x7FFFFFFF;
const long long nmos = 0x80000000;
const int inf = 0x3f3f3f3f;
const long long inff = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e8 + 7;
const double esp = 1e-8;
const double PI = acos(-1.0);
const double PHI = 0.61803399;
const double tPHI = 0.38196601;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
template <typename T>
inline void mod_(T &A, long long MOD = mod) {
  A %= MOD;
  A += MOD;
  A %= MOD;
}
template <typename T>
inline T read(T &x) {
  x = 0;
  int f = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') f |= (ch == '-'), ch = getchar();
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  return x = f ? -x : x;
}
int main() {
  int T, n;
  scanf("%d", &T);
  while (T--) {
    scanf("%d", &n);
    if (n * n - 4 * n < 0) {
      puts("N");
      continue;
    }
    double d = n * 1.0;
    double ans = d * 1.0 + sqrt(1.0 * d * d - 4.0 * d);
    ans = ans / 2.0;
    printf("Y %.9f %.9f\n", ans, 1.0 * d - ans);
  }
  return 0;
}
