Problem: Try guessing the statement from this picture: 

<image>

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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const long long inf = 0x3f3f3f3f;
const int N = 1e6 + 10;
const long long mod = 998244353;
map<string, long long> mp;
map<string, int> ml;
int c[N], vis[N], a[N], t, n, m, x, y, ans;
char s[N];
int ex, ey, cnt;
long long p(long long x) {
  for (long long i = 2; i <= sqrt(x); i++) {
    if (n % i == 0) return 0;
  }
  return 1;
}
long long f(long long x) {
  for (long long i = 2; i <= n; i++) {
    if (n % i == 0 && p(i)) return i;
  }
}
int main() {
  scanf("%d", &n);
  while (n--) {
    double m;
    scanf("%lf", &m);
    double l, r;
    if (m == 0) {
      cout << "Y 0.000000000 0.000000000" << endl;
    } else {
      l = 0;
      r = m;
      double mid;
      for (int i = 1; i <= 50; i++) {
        mid = (l + r) / 2;
        double t = mid * mid - m * mid + m;
        if (t > 0) {
          r = mid - 1;
        } else
          l = mid + 1;
      }
      if (mid > 0 && m - mid > 0)
        printf("Y %.9f %.9f\n", mid, m - mid);
      else
        cout << "N" << endl;
    }
  }
}