#include <bits/stdc++.h>
using namespace std;
template <class X, class Y>
void minimize(X &x, const Y &y) {
  if (x > y) {
    x = y;
  }
}
template <class X, class Y>
void maximize(X &x, const Y &y) {
  if (x < y) {
    x = y;
  }
}
long long x, y, m;
void init() { cin >> x >> y >> m; }
void solve() {
  if (x <= 0 && y <= 0 && m > 0) {
    cout << -1 << "\n";
    return;
  }
  if (max(x, y) >= m) {
    cout << 0 << "\n";
    return;
  }
  long long cou = 0, step;
  if (x < 0) {
    step = abs(x) / y;
    if (abs(x) % y) step++;
    x += y * step;
    cou = step;
  } else if (y < 0) {
    step = abs(y) / x;
    if (abs(y) % x) step++;
    y += x * step;
    cou = step;
  }
  long long tmp;
  while (max(x, y) < m) {
    tmp = x + y;
    if (x < y)
      x = x + y;
    else
      y = x + y;
    if (x + y < tmp) {
      cout << -1 << "\n";
      return;
    }
    cou++;
  }
  cout << cou << "\n";
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  init();
  solve();
}

### Prompt

Develop a solution in Cpp to the problem described below:

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 X, class Y>
void minimize(X &x, const Y &y) {
  if (x > y) {
    x = y;
  }
}
template <class X, class Y>
void maximize(X &x, const Y &y) {
  if (x < y) {
    x = y;
  }
}
long long x, y, m;
void init() { cin >> x >> y >> m; }
void solve() {
  if (x <= 0 && y <= 0 && m > 0) {
    cout << -1 << "\n";
    return;
  }
  if (max(x, y) >= m) {
    cout << 0 << "\n";
    return;
  }
  long long cou = 0, step;
  if (x < 0) {
    step = abs(x) / y;
    if (abs(x) % y) step++;
    x += y * step;
    cou = step;
  } else if (y < 0) {
    step = abs(y) / x;
    if (abs(y) % x) step++;
    y += x * step;
    cou = step;
  }
  long long tmp;
  while (max(x, y) < m) {
    tmp = x + y;
    if (x < y)
      x = x + y;
    else
      y = x + y;
    if (x + y < tmp) {
      cout << -1 << "\n";
      return;
    }
    cou++;
  }
  cout << cou << "\n";
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  init();
  solve();
}

```