很簡單的一元二次方程，用高中學的公式求解就好啦。 

大晚上打比賽真刺激！！！嘿嘿嘿

C. Meme Problem

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Try guessing the statement from this picture:

You are given a non-negative integer dd. You have to find two non-negative real numbers aa and bb such that a+b=da+b=d and a⋅b=da⋅b=d.

Input

The first line contains tt (1≤t≤1031≤t≤103) — the number of test cases.

Each test case contains one integer dd (0≤d≤103)(0≤d≤103).

Output

For each test print one line.

If there is an answer for the ii-th test, print "Y", and then the numbers aa and bb.

If there is no answer for the ii-th test, print "N".

Your answer will be considered correct if |(a+b)−a⋅b|≤10−6|(a+b)−a⋅b|≤10−6 and |(a+b)−d|≤10−6|(a+b)−d|≤10−6.

Example

input

Copy

7
69
0
1
4
5
999
1000

output

Copy

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<iostream>
#include<cmath>
#include<cstdio>
using namespace std;

int main()
{
    int n,m,j,k,i,T;
    double a;
    cin>>T;
    while (T--)
    {
        cin>>a;
        if (a*a-4*a<0)
            printf("N\n");
        else
        {
            double ans = (a+sqrt(a*a-4*a))/2.0;
            printf("Y %.9lf %.9lf\n",ans,a-ans);
        }
    }
    return 0;
}