#include <bits/stdc++.h>
using namespace std;
template <class T>
string str(T Number) {
  string Result;
  ostringstream convert;
  convert << Number;
  Result = convert.str();
  return Result;
}
int StringToNumber(const string &Text) {
  istringstream ss(Text);
  int result;
  return ss >> result ? result : 0;
}
template <class T>
inline vector<pair<T, int> > FACTORISE(T n) {
  vector<pair<T, int> > R;
  for (T i = 2; n > 1;) {
    if (n % i == 0) {
      int C = 0;
      for (; n % i == 0; C++, n /= i)
        ;
      R.push_back(make_pair(i, C));
    }
    i++;
    if (i > n / i) i = n;
  }
  if (n > 1) R.push_back(make_pair(n, 1));
  return R;
}
template <class T>
inline T TOTIENT(T n) {
  vector<pair<T, int> > R = FACTORISE(n);
  T r = n;
  for (int i = 0; i < R.size(); i++) r = r / R[i].first * (R[i].first - 1);
  return r;
}
double rnd(float d) { return floor(d + 0.49); }
int main() {
  long long x, y, m;
  cin >> x >> y >> m;
  long long count = 0;
  if (y < x) swap(x, y);
  if (m <= y) {
    cout << 0 << endl;
    return 0;
  }
  if (y <= 0) {
    cout << -1 << endl;
    return 0;
  }
  if (x < 0) {
    count += abs(x) / y;
    x += (count * y);
    while (x <= 0) {
      x += y;
      count++;
    }
  }
  while (y < m) {
    x += y;
    swap(x, y);
    count++;
  }
  cout << count << endl;
  return 0;
}

### Prompt

Generate a CPP solution to the following problem:

Let us call a pair of integer numbers m-perfect, if at least one number in the pair is greater than or equal to m. Thus, the pairs (3, 3) and (0, 2) are 2-perfect while the pair (-1, 1) is not.

Two integers x, y are written on the blackboard. It is allowed to erase one of them and replace it with the sum of the numbers, (x + y).

What is the minimum number of such operations one has to perform in order to make the given pair of integers m-perfect?

Input

Single line of the input contains three integers x, y and m ( - 1018 ≤ x, y, m ≤ 1018).

Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preffered to use the cin, cout streams or the %I64d specifier.

Output

Print the minimum number of operations or "-1" (without quotes), if it is impossible to transform the given pair to the m-perfect one.

Examples

Input

1 2 5


Output

2


Input

-1 4 15


Output

4


Input

0 -1 5


Output

-1

Note

In the first sample the following sequence of operations is suitable: (1, 2) <image> (3, 2) <image> (5, 2).

In the second sample: (-1, 4) <image> (3, 4) <image> (7, 4) <image> (11, 4) <image> (15, 4).

Finally, in the third sample x, y cannot be made positive, hence there is no proper sequence of operations.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
template <class T>
string str(T Number) {
  string Result;
  ostringstream convert;
  convert << Number;
  Result = convert.str();
  return Result;
}
int StringToNumber(const string &Text) {
  istringstream ss(Text);
  int result;
  return ss >> result ? result : 0;
}
template <class T>
inline vector<pair<T, int> > FACTORISE(T n) {
  vector<pair<T, int> > R;
  for (T i = 2; n > 1;) {
    if (n % i == 0) {
      int C = 0;
      for (; n % i == 0; C++, n /= i)
        ;
      R.push_back(make_pair(i, C));
    }
    i++;
    if (i > n / i) i = n;
  }
  if (n > 1) R.push_back(make_pair(n, 1));
  return R;
}
template <class T>
inline T TOTIENT(T n) {
  vector<pair<T, int> > R = FACTORISE(n);
  T r = n;
  for (int i = 0; i < R.size(); i++) r = r / R[i].first * (R[i].first - 1);
  return r;
}
double rnd(float d) { return floor(d + 0.49); }
int main() {
  long long x, y, m;
  cin >> x >> y >> m;
  long long count = 0;
  if (y < x) swap(x, y);
  if (m <= y) {
    cout << 0 << endl;
    return 0;
  }
  if (y <= 0) {
    cout << -1 << endl;
    return 0;
  }
  if (x < 0) {
    count += abs(x) / y;
    x += (count * y);
    while (x <= 0) {
      x += y;
      count++;
    }
  }
  while (y < m) {
    x += y;
    swap(x, y);
    count++;
  }
  cout << count << endl;
  return 0;
}

```