#include <bits/stdc++.h>
using namespace std;
const int MAXN = 2e5 + 5;
const int INF = 2e9;
const int MOD = 1e9 + 7;
long long dp[MAXN][2];
int main() {
  long long A = 0, B = 0, edge = 0, cnt = 0;
  scanf("%I64d%I64d%I64d", &A, &B, &edge);
  if (edge <= A || edge <= B) return printf("0"), 0;
  if (A <= 0 && B <= 0 && (edge > A && edge > B)) return printf("-1"), 0;
  while (A < edge && B < edge) {
    if (A < B) {
      long long ways = (B - A) / B + 1;
      A += B * ways;
      cnt += ways;
    } else {
      long long ways = (A - B) / A + 1;
      B += A * ways;
      cnt += ways;
    }
  }
  printf("%I64d", cnt);
  return 0;
}

### 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;
const int MAXN = 2e5 + 5;
const int INF = 2e9;
const int MOD = 1e9 + 7;
long long dp[MAXN][2];
int main() {
  long long A = 0, B = 0, edge = 0, cnt = 0;
  scanf("%I64d%I64d%I64d", &A, &B, &edge);
  if (edge <= A || edge <= B) return printf("0"), 0;
  if (A <= 0 && B <= 0 && (edge > A && edge > B)) return printf("-1"), 0;
  while (A < edge && B < edge) {
    if (A < B) {
      long long ways = (B - A) / B + 1;
      A += B * ways;
      cnt += ways;
    } else {
      long long ways = (A - B) / A + 1;
      B += A * ways;
      cnt += ways;
    }
  }
  printf("%I64d", cnt);
  return 0;
}

```