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
#include <bits/stdc++.h>
using namespace std;
bool SR(int &x) { return scanf("%d", &x) == 1; }
bool SR(long long &x) { return scanf("%lld", &x) == 1; }
bool SR(double &x) { return scanf("%lf", &x) == 1; }
bool SR(char *s) { return scanf("%s", s) == 1; }
bool RI() { return true; }
template <typename I, typename... T>
bool RI(I &x, T &...tail) {
  return SR(x) && RI(tail...);
}
void SP(const int x) { printf("%d", x); }
void SP(const long long x) { printf("%lld", x); }
void SP(const double x) { printf("%.16lf", x); }
void SP(const char *s) { printf("%s", s); }
void PL() { puts(""); }
template <typename I, typename... T>
void PL(const I x, const T... tail) {
  SP(x);
  if (sizeof...(tail)) putchar(' ');
  PL(tail...);
}
int ar[3];
void read() {
  for (int i = 0; i < int(3); i++) RI(ar[i]);
}
void build() {}
void sol0(int a) {
  if (a == 0)
    PL(-1);
  else
    PL(0);
}
void sol1(int a, int b) {
  PL(1);
  PL((0.0 - b) / a);
}
void sol2(long long a, long long b, long long c) {
  long long d = b * b - 4 * a * c;
  if (d < 0)
    PL(0);
  else if (d == 0) {
    PL(1);
    PL((0.0 - b) / (2 * a));
  } else {
    double p = sqrt(d);
    vector<double> v;
    v.push_back((-b + p) / (2 * a));
    v.push_back((-b - p) / (2 * a));
    sort(begin(v), end(v));
    PL(2);
    for (auto x : v) PL(x);
  }
}
void sol() {
  if (ar[0] == 0 && ar[1] == 0)
    sol0(ar[2]);
  else if (ar[0] == 0)
    sol1(ar[1], ar[2]);
  else
    sol2(ar[0], ar[1], ar[2]);
}
int main() {
  read();
  build();
  sol();
  return 0;
}
