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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 double eps = 1e-10;
const double whs = 1e-10;
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    int d;
    scanf("%d", &d);
    double sta = d;
    double l = 0;
    double r = sta / 2;
    int flag = 0;
    double ans;
    while (l < r) {
      double mid = (l + r) / 2;
      double res = (sta - mid) * mid;
      if (res - sta <= eps && res - sta >= -eps) {
        ans = mid;
        flag = 1;
        break;
      } else if (res > sta + eps)
        r = mid;
      else
        l = mid;
    }
    double res = (sta - l) * l;
    if (res - sta <= eps && res - sta >= -eps) {
      ans = l;
      flag = 1;
    }
    if (flag) {
      printf("Y %.9lf %.9lf\n", ans, sta - ans);
    } else
      printf("N\n");
  }
  return 0;
}
