#include <bits/stdc++.h>
using namespace std;
long long i, j, k, N, M, A[100000], x, y, a, b, aa, bb;
bool outrange(long long x, long long y) {
  if (x <= j && y <= j && x >= -1 * j && y >= -1 * j) return true;
  return false;
}
int main() {
  j = 1;
  for (i = 1; i <= 62; i++) j *= 2;
  j = j * 2 - 1;
  scanf("%I64d%I64d%I64d", &x, &y, &M);
  if (x >= M || y >= M) {
    printf("0\n");
    return 0;
  } else if (x <= 0 && y <= 0) {
    printf("-1\n");
    return 0;
  }
  aa = x, bb = y;
  if (aa > bb) {
    swap(aa, bb);
    swap(x, y);
  }
  i = 0;
  if (aa < 0) {
    long long xx = -(aa) / bb;
    aa = aa - ((aa) / bb) * bb + bb;
    i = xx + 1;
  }
  x = aa;
  for (i = i + 1;; i++) {
    if (x > y)
      y = x + y;
    else
      x = x + y;
    if (x >= M || y >= M) {
      printf("%I64d\n", i);
      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;
long long i, j, k, N, M, A[100000], x, y, a, b, aa, bb;
bool outrange(long long x, long long y) {
  if (x <= j && y <= j && x >= -1 * j && y >= -1 * j) return true;
  return false;
}
int main() {
  j = 1;
  for (i = 1; i <= 62; i++) j *= 2;
  j = j * 2 - 1;
  scanf("%I64d%I64d%I64d", &x, &y, &M);
  if (x >= M || y >= M) {
    printf("0\n");
    return 0;
  } else if (x <= 0 && y <= 0) {
    printf("-1\n");
    return 0;
  }
  aa = x, bb = y;
  if (aa > bb) {
    swap(aa, bb);
    swap(x, y);
  }
  i = 0;
  if (aa < 0) {
    long long xx = -(aa) / bb;
    aa = aa - ((aa) / bb) * bb + bb;
    i = xx + 1;
  }
  x = aa;
  for (i = i + 1;; i++) {
    if (x > y)
      y = x + y;
    else
      x = x + y;
    if (x >= M || y >= M) {
      printf("%I64d\n", i);
      return 0;
    }
  }
}

```