#include <bits/stdc++.h>
using namespace std;
long long x, y, m, sum, unk;
long long func(long long a, long long b, long long c) {
  while (1) {
    if (a >= c || b >= c) {
      cout << sum;
      break;
    } else if (a >= b)
      b = a + b;
    else
      a = a + b;
    sum++;
  }
  return sum;
}
int main() {
  cin >> x >> y >> m;
  if (x >= m || y >= m) {
    cout << 0;
    return 0;
  }
  if ((x == 0 && y < 0) || (x < 0 && y == 0) || (x < 0 && y < 0 && x + y < m) ||
      (x >= 0 && y >= 0 && (x > m || y > m)) || (x <= 0 && y <= 0 && m > 0)) {
    cout << "-1";
    return 0;
  } else {
    if (x >= 0 && y >= 0)
      func(x, y, m);
    else {
      if (x < 0 && y >= 0) {
        unk = abs(x % y);
        long long z = x;
        x += unk;
        sum = abs(x / y);
        if (z % y == 0)
          x = 0;
        else {
          x = z % y + y;
          sum++;
        }
        func(x, y, m);
      } else if (y < 0 && x >= 0) {
        unk = abs(y % x);
        long long z = y;
        y += unk;
        sum = abs(y / x);
        if (z % x == 0)
          y = 0;
        else {
          y = z % x + x;
          sum++;
        }
        func(x, y, m);
      } else
        func(abs(x), abs(y), abs(m));
    }
  }
}

### Prompt

Your challenge is to write 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;
long long x, y, m, sum, unk;
long long func(long long a, long long b, long long c) {
  while (1) {
    if (a >= c || b >= c) {
      cout << sum;
      break;
    } else if (a >= b)
      b = a + b;
    else
      a = a + b;
    sum++;
  }
  return sum;
}
int main() {
  cin >> x >> y >> m;
  if (x >= m || y >= m) {
    cout << 0;
    return 0;
  }
  if ((x == 0 && y < 0) || (x < 0 && y == 0) || (x < 0 && y < 0 && x + y < m) ||
      (x >= 0 && y >= 0 && (x > m || y > m)) || (x <= 0 && y <= 0 && m > 0)) {
    cout << "-1";
    return 0;
  } else {
    if (x >= 0 && y >= 0)
      func(x, y, m);
    else {
      if (x < 0 && y >= 0) {
        unk = abs(x % y);
        long long z = x;
        x += unk;
        sum = abs(x / y);
        if (z % y == 0)
          x = 0;
        else {
          x = z % y + y;
          sum++;
        }
        func(x, y, m);
      } else if (y < 0 && x >= 0) {
        unk = abs(y % x);
        long long z = y;
        y += unk;
        sum = abs(y / x);
        if (z % x == 0)
          y = 0;
        else {
          y = z % x + x;
          sum++;
        }
        func(x, y, m);
      } else
        func(abs(x), abs(y), abs(m));
    }
  }
}

```