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;
char BUFFER[100000 + 5];
bool readn(int &n) { return scanf("%d", &n) == 1; }
bool readl(long long &n) { return scanf("%I64d", &n) == 1; }
bool readd(double &n) { return scanf("%lf", &n) == 1; }
bool reads(string &s) {
  s = "";
  int n = scanf("%s", BUFFER);
  if (n == 1) s = BUFFER;
  return n == 1;
}
bool readln(string &s) {
  char *valid = gets(BUFFER);
  if (valid) s = BUFFER;
  return ((bool)valid);
}
void Print(long long &n, bool printNewLine = true) {
  printNewLine ? printf("%I64d\n", n) : printf("%I64d", n);
}
long long calc() {
  long long a, b, m, cnt = 0;
  cin >> a >> b >> m;
  if (a > b) swap(a, b);
  if (b >= m) return 0;
  if (b <= 0) return -1;
  if (a < 0) {
    a = -a;
    cnt = a / b;
    a %= b;
    a = -a;
  }
  while (1) {
    if (a > b) swap(a, b);
    if (b >= m) return cnt;
    a += b;
    cnt++;
  }
}
int main() {
  cout << calc() << endl;
  return 0;
}
