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.
#include <bits/stdc++.h>
using namespace std;
template <typename T>
string toString(T x) {
  if (x == 0) return "0";
  bool negative = x < 0;
  string res;
  while (x) {
    res.push_back('0' + x % 10);
    x /= 10;
  }
  if (negative) res.push_back('-');
  reverse(res.begin(), res.end());
  return res;
}
void ioFromFile(string inputFileName, string outputFileName) {
  ifstream fin(inputFileName);
  ofstream fout(outputFileName);
  cin.rdbuf(fin.rdbuf());
  cout.rdbuf(fout.rdbuf());
}
int read() {
  int res;
  scanf("%d", &res);
  return res;
}
void write(int x) { printf("%d ", x); }
template <typename T>
T gcd(T a, T b) {
  return a == 0 ? b : gcd(b % a, a);
}
template <typename T>
T lcm(T a, T b) {
  return a / gcd(a, b) * b;
}
int main() {
  long long a, b, k;
  cin >> a >> b >> k;
  if (a > b) swap(a, b);
  if (b >= k) {
    cout << 0;
    return 0;
  }
  if (b <= 0) {
    cout << -1;
    return 0;
  }
  long long res = 0;
  if (a < 0) {
    res = (-a) / b + 1;
    a += res * b;
  }
  while (b < k) {
    res++;
    a += b;
    swap(a, b);
  }
  cout << res;
  return 0;
}
