Problem: You are given an equation: 

Ax2 + 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 ( - 105 ≤ A, B, C ≤ 105). 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const long double EPS = (long double)1e-7;
const long double PI = acos(0) * 2;
bool isZero(const long double& x) { return abs(x) <= EPS; }
int sign(const long double& x) { return isZero(x) ? 0 : (0 < x ? 1 : -1); }
long long gcd(long long a, long long b) {
  for (; b; a %= b, swap(a, b)) {
  }
  return abs(a);
}
pair<long long, long long> operator+(const pair<long long, long long>& a,
                                     const pair<long long, long long>& b) {
  return pair<long long, long long>(a.first + b.first, a.second + b.second);
}
pair<long long, long long> operator-(const pair<long long, long long>& a,
                                     const pair<long long, long long>& b) {
  return pair<long long, long long>(a.first - b.first, a.second - b.second);
}
pair<long long, long long> operator*(const pair<long long, long long>& a,
                                     const long long& b) {
  return pair<long long, long long>(a.first * b, a.second * b);
}
long long operator*(const pair<long long, long long>& a,
                    const pair<long long, long long>& b) {
  return a.first * b.second - b.first * a.second;
}
long long ccw(const pair<long long, long long>& a,
              const pair<long long, long long>& b,
              const pair<long long, long long>& c) {
  return a * b + b * c + c * a;
}
void fg(vector<int> G[], int a, int b) {
  G[a].emplace_back(b);
  G[b].emplace_back(a);
}
void fg(vector<pair<int, int> > G[], int a, int b, int c) {
  G[a].emplace_back(b, c);
  G[b].emplace_back(a, c);
}
long long A, B, C;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> A >> B >> C;
  if (!A && !B) {
    puts(C ? "0" : "-1");
    exit(0);
  }
  if (!A) {
    printf("1\n%.10lf\n", double(-C) / B);
    exit(0);
  }
  long long D = B * B - 4 * A * C;
  if (D < 0) {
    puts("0");
    exit(0);
  }
  if (!D) {
    printf("1\n%.10lf\n", double(-B) / A / 2);
    exit(0);
  }
  if (A < 0) {
    A = -A;
    B = -B;
    C = -C;
  }
  printf("2\n%.10lf\n%.10lf\n", double(-B - sqrt(D)) / A / 2,
         double(-B + sqrt(D)) / A / 2);
  return 0;
}