#include <bits/stdc++.h>
using namespace std;
long long x, y, m;
long long process1() {
  long long fa[111], fb[111];
  fa[0] = 1;
  fa[1] = 1;
  fb[0] = 0;
  fb[1] = 1;
  int res = 0;
  long long sum = x;
  int idx = 1;
  while (sum < m) {
    res++;
    sum = fa[idx] * x + fb[idx] * y;
    idx++;
    if (idx > 1) {
      fa[idx] = fa[idx - 1] + fa[idx - 2];
      fb[idx] = fb[idx - 1] + fb[idx - 2];
    }
  }
  return res;
}
long long process2() {
  long long k = -y;
  long long tmp = 0;
  tmp = k / x;
  if (k % x != 0) tmp++;
  y += tmp * x;
  if (x < y) swap(x, y);
  return tmp + process1();
}
int main() {
  cin >> x >> y >> m;
  if (x < y) swap(x, y);
  if (x >= m) {
    printf("0\n");
    return 0;
  }
  if (x == 0) {
    if (x >= m)
      printf("0\n");
    else
      printf("-1\n");
    return 0;
  }
  if (x > 0 && y >= 0) {
    cout << process1() << endl;
    return 0;
  }
  if (x > 0 && y < 0) {
    cout << process2() << endl;
    return 0;
  }
  if (x < 0 && y < 0) {
    if (x > m || y > m)
      printf("0\n");
    else
      printf("-1\n");
  }
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;
long long process1() {
  long long fa[111], fb[111];
  fa[0] = 1;
  fa[1] = 1;
  fb[0] = 0;
  fb[1] = 1;
  int res = 0;
  long long sum = x;
  int idx = 1;
  while (sum < m) {
    res++;
    sum = fa[idx] * x + fb[idx] * y;
    idx++;
    if (idx > 1) {
      fa[idx] = fa[idx - 1] + fa[idx - 2];
      fb[idx] = fb[idx - 1] + fb[idx - 2];
    }
  }
  return res;
}
long long process2() {
  long long k = -y;
  long long tmp = 0;
  tmp = k / x;
  if (k % x != 0) tmp++;
  y += tmp * x;
  if (x < y) swap(x, y);
  return tmp + process1();
}
int main() {
  cin >> x >> y >> m;
  if (x < y) swap(x, y);
  if (x >= m) {
    printf("0\n");
    return 0;
  }
  if (x == 0) {
    if (x >= m)
      printf("0\n");
    else
      printf("-1\n");
    return 0;
  }
  if (x > 0 && y >= 0) {
    cout << process1() << endl;
    return 0;
  }
  if (x > 0 && y < 0) {
    cout << process2() << endl;
    return 0;
  }
  if (x < 0 && y < 0) {
    if (x > m || y > m)
      printf("0\n");
    else
      printf("-1\n");
  }
  return 0;
}

```