20B. Equation

math, 2000, https://codeforces.com/problemset/problem/20/B

You are given an equation:

Ax^2 + Bx + C = 0.

Your task is to find the number of distinct roots of the equation and print all of them in ascending order.

Input

The first line contains three integer numbers A, B and C ( - 10^5 ≤ A, B, C ≤ 10^5). Any coefficient may be equal to 0.

Output

In case of infinite root count print the only integer -1. In case of no roots print the only integer 0. In other cases print the number of root on the first line and the roots on the following lines in the ascending order. Print roots with at least 5 digits after the decimal point.

Examples

input

1 -5 6

output

2
2.0000000000
3.0000000000

分情况处理：

A = 0 → 方程退化为线性方程或常数方程：

如果 B = 0, C = 0 → 恒等式 0 = 0，解无限多 → 输出 -1。
如果 B = 0, C ≠ 0 → 恒不成立 → 输出 0。
否则是线性方程 Bx + C = 0 → 唯一解 x = -C/B。

A ≠ 0 → 标准二次方程，计算判别式 D = B^2 - 4AC：

如果 D < 0 → 无解 → 输出 0。
如果 D = 0 → 一个实根 -B / (2A)。
如果 D > 0 → 两个实根，按升序输出。

输出时要保证小数点后至少 5 位（用 :.10f 来格式化即可）。

python

import math

def solve():
    A, B, C = map(int, input().split())
    
    # Case 1: Linear / constant equation
    if A == 0:
        if B == 0:
            if C == 0:
                print(-1)  # infinite solutions
            else:
                print(0)   # no solution
        else:
            root = -C / B
            print(1)
            print(f"{root:.10f}")
        return
    
    # Case 2: Quadratic equation
    D = B * B - 4 * A * C
    if D < 0:
        print(0)
    elif D == 0:
        root = -B / (2 * A)
        print(1)
        print(f"{root:.10f}")
    else:
        sqrt_D = math.sqrt(D)
        r1 = (-B - sqrt_D) / (2 * A)
        r2 = (-B + sqrt_D) / (2 * A)
        if r1 > r2:
            r1, r2 = r2, r1
        print(2)
        print(f"{r1:.10f}")
        print(f"{r2:.10f}")

if __name__ == "__main__":
    solve()

20B. Equation ​

20B. Equation