#include <bits/stdc++.h>
int nextInt() {
  int x;
  scanf("%d", &x);
  return x;
}
double nextDouble() {
  double x;
  scanf("%lf", &x);
  return x;
}
long long nextLong() {
  long long x;
  scanf("%I64d", &x);
  return x;
}
char nextChar() {
  char x;
  scanf("%c", &x);
  return x;
}
void newLine() { printf("\n"); }
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
long long pow(long long a, long long b, long long MOD) {
  long long x = 1, y = a;
  while (b > 0) {
    if (b % 2 == 1) {
      x = (x * y);
      if (x > MOD) x %= MOD;
    }
    y = (y * y);
    if (y > MOD) y %= MOD;
    b /= 2;
  }
  return x;
}
int countSetBit(long long n) {
  int ans = 0;
  while (n != 0) {
    n -= (n & -n);
    ++ans;
  }
  return ans;
}
long long mod = 1e9 + 7;
const int N = 30;
const int M = 55;
const double eps = 1e-6;
using namespace std;
int dp[N][N] = {0};
int g[N][N];
int main() {
  double a, b, c;
  cin >> a >> b >> c;
  cout << fixed << setprecision(10);
  if (a == 0 && b == 0 && c == 0) {
    cout << -1 << endl;
    return 0;
  }
  if (a == 0 && b == 0 && c != 0) {
    cout << 0 << endl;
    return 0;
  }
  if (a == 0 && b != 0) {
    cout << "1\n" << -c / b << endl;
    return 0;
  }
  if (a != 0 && b == 0) {
    if (-1.0 * c / a == 0)
      cout << "1\n0.000000" << endl;
    else if (-1.0 * c / a < 0)
      cout << 0 << endl;
    else
      cout << "2\n"
           << sqrt(-1.0 * c / a) << "\n"
           << -1.0 * sqrt(-1.0 * c / a) << endl;
    return 0;
  }
  double delta = b * b - 4 * a * c;
  if (delta < 0) {
    cout << 0 << endl;
    return 0;
  }
  double sq = sqrt(delta);
  double x1 = (-b - sq) / (2 * a);
  double x2 = (-b + sq) / (2 * a);
  if (x1 - x2 > eps) swap(x1, x2);
  if (delta == 0)
    cout << "1\n" << x1 << endl;
  else {
    cout << "2\n" << x1 << "\n" << x2 << endl;
  }
}

### Prompt

Please formulate a CPP solution to 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>
int nextInt() {
  int x;
  scanf("%d", &x);
  return x;
}
double nextDouble() {
  double x;
  scanf("%lf", &x);
  return x;
}
long long nextLong() {
  long long x;
  scanf("%I64d", &x);
  return x;
}
char nextChar() {
  char x;
  scanf("%c", &x);
  return x;
}
void newLine() { printf("\n"); }
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
long long pow(long long a, long long b, long long MOD) {
  long long x = 1, y = a;
  while (b > 0) {
    if (b % 2 == 1) {
      x = (x * y);
      if (x > MOD) x %= MOD;
    }
    y = (y * y);
    if (y > MOD) y %= MOD;
    b /= 2;
  }
  return x;
}
int countSetBit(long long n) {
  int ans = 0;
  while (n != 0) {
    n -= (n & -n);
    ++ans;
  }
  return ans;
}
long long mod = 1e9 + 7;
const int N = 30;
const int M = 55;
const double eps = 1e-6;
using namespace std;
int dp[N][N] = {0};
int g[N][N];
int main() {
  double a, b, c;
  cin >> a >> b >> c;
  cout << fixed << setprecision(10);
  if (a == 0 && b == 0 && c == 0) {
    cout << -1 << endl;
    return 0;
  }
  if (a == 0 && b == 0 && c != 0) {
    cout << 0 << endl;
    return 0;
  }
  if (a == 0 && b != 0) {
    cout << "1\n" << -c / b << endl;
    return 0;
  }
  if (a != 0 && b == 0) {
    if (-1.0 * c / a == 0)
      cout << "1\n0.000000" << endl;
    else if (-1.0 * c / a < 0)
      cout << 0 << endl;
    else
      cout << "2\n"
           << sqrt(-1.0 * c / a) << "\n"
           << -1.0 * sqrt(-1.0 * c / a) << endl;
    return 0;
  }
  double delta = b * b - 4 * a * c;
  if (delta < 0) {
    cout << 0 << endl;
    return 0;
  }
  double sq = sqrt(delta);
  double x1 = (-b - sq) / (2 * a);
  double x2 = (-b + sq) / (2 * a);
  if (x1 - x2 > eps) swap(x1, x2);
  if (delta == 0)
    cout << "1\n" << x1 << endl;
  else {
    cout << "2\n" << x1 << "\n" << x2 << endl;
  }
}

```