Consider an N-degree polynomial, expressed as follows:

**P**N * xN \+ **P**N-1 * xN-1 \+ ... + **P**1 * x1 \+ **P**0 * x0

You'd like to find all of the polynomial's x-intercepts — in other words, all
distinct real values of x for which the expression evaluates to 0.

Unfortunately, the order of operations has been reversed: Addition (**+**) now
has the highest precedence, followed by multiplication (*****), followed by
exponentiation (**^**). In other words, an expression like ab \+ c * d should
be evaluated as a((b+c)*d). For our purposes, exponentiation is right-
associative (in other words, abc = a(bc)), and 00 = 1. The unary negation
operator still has the highest precedence, so the expression -2-3 * -1 + -2
evaluates to -2(-3 * (-1 + -2)) = -29 = -512.

### Input

Input begins with an integer **T**, the number of polynomials. For each
polynomial, there is first a line containing the integer **N**, the degree of
the polynomial. Then, **N**+1 lines follow. The _i_th of these lines contains
the integer **Pi-1**.

### Output

For the _i_th polynomial, print a line containing "Case #_i_: **K**", where
**K** is the number of distinct real values of **x** for which the polynomial
evaluates to 0. Then print **K** lines, each containing such a value of **x**,
in increasing order.

Absolute and relative errors of up to 10-6 will be ignored in the x-intercepts
you output. However, **K** must be exactly correct.

### Constraints

1 ≤ **T** ≤ 200  
0 ≤ **N** ≤ 50  
-50 ≤ **Pi** ≤ 50   
**PN** ≠ 0   

### Explanation of Sample

In the first case, the polynomial is 1 * x1 \+ 1 * x0. With the order of
operations reversed, this is evaluated as (1 * x)(((1 + 1) * x)0), which is
equal to 0 only when x = 0.

In the second case, the polynomial does not evaluate to 0 for any real value
x.