#include <bits/stdc++.h>
using namespace std;
void SieveOfEratosthenes(long long n) {
  bool prime[n + 1];
  memset(prime, true, sizeof(prime));
  for (long long p = 2; p * p <= n; p++) {
    if (prime[p] == true) {
      for (long long i = p * p; i <= n; i += p) prime[i] = false;
    }
  }
}
long long power(long long x, long long y, long long m) {
  if (y == 0) return 1;
  long long p = power(x, y / 2, m) % m;
  p = (p * p) % m;
  return (y % 2 == 0) ? p : (x * p) % m;
}
void print(vector<long long> a) {
  for (long long i = 0; i < a.size(); i++) cout << a[i] << " ";
  cout << "\n";
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return (a / gcd(a, b)) * b; }
bool cmp(long long a, long long b) { return a > b; }
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long a, b, c;
  cin >> a >> b >> c;
  if (a) {
    if (b) {
      long long d = b * b - 4 * a * c;
      if (d >= 0) {
        double f = sqrt(d);
        double r1 = -b + f;
        r1 = double(r1) / double(2 * a);
        double r2 = -b - f;
        r2 = double(r2) / double(2 * a);
        if (r1 != r2) {
          if (r2 > r1) swap(r1, r2);
          cout << 2 << "\n" << fixed << setprecision(8) << r2 << "\n" << r1;
        } else
          cout << 1 << "\n" << fixed << setprecision(8) << r2;
      } else {
        cout << 0;
      }
    } else {
      double d = -c;
      d = d / double(a);
      if (d >= 0) {
        cout << 1 << "\n" << fixed << setprecision(8) << sqrt(d);
      } else {
        cout << 0;
      }
    }
  } else {
    if (b) {
      double d = double(-c) / double(b);
      cout << 1 << "\n" << fixed << setprecision(8) << d;
    } else {
      if (c)
        cout << 0;
      else
        cout << -1;
    }
  }
}

### Prompt

Create a solution in CPP for the following 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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
void SieveOfEratosthenes(long long n) {
  bool prime[n + 1];
  memset(prime, true, sizeof(prime));
  for (long long p = 2; p * p <= n; p++) {
    if (prime[p] == true) {
      for (long long i = p * p; i <= n; i += p) prime[i] = false;
    }
  }
}
long long power(long long x, long long y, long long m) {
  if (y == 0) return 1;
  long long p = power(x, y / 2, m) % m;
  p = (p * p) % m;
  return (y % 2 == 0) ? p : (x * p) % m;
}
void print(vector<long long> a) {
  for (long long i = 0; i < a.size(); i++) cout << a[i] << " ";
  cout << "\n";
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return (a / gcd(a, b)) * b; }
bool cmp(long long a, long long b) { return a > b; }
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long a, b, c;
  cin >> a >> b >> c;
  if (a) {
    if (b) {
      long long d = b * b - 4 * a * c;
      if (d >= 0) {
        double f = sqrt(d);
        double r1 = -b + f;
        r1 = double(r1) / double(2 * a);
        double r2 = -b - f;
        r2 = double(r2) / double(2 * a);
        if (r1 != r2) {
          if (r2 > r1) swap(r1, r2);
          cout << 2 << "\n" << fixed << setprecision(8) << r2 << "\n" << r1;
        } else
          cout << 1 << "\n" << fixed << setprecision(8) << r2;
      } else {
        cout << 0;
      }
    } else {
      double d = -c;
      d = d / double(a);
      if (d >= 0) {
        cout << 1 << "\n" << fixed << setprecision(8) << sqrt(d);
      } else {
        cout << 0;
      }
    }
  } else {
    if (b) {
      double d = double(-c) / double(b);
      cout << 1 << "\n" << fixed << setprecision(8) << d;
    } else {
      if (c)
        cout << 0;
      else
        cout << -1;
    }
  }
}

```