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;
const int maxn = 1e18 + 3;
const int maxm = 1e6 + 7;
long long x, y, m;
long long dp[maxm];
long long ans;
void solve() {
  dp[0] = min(x, y);
  dp[1] = max(x, y);
  long long a = dp[0];
  long long b = dp[1];
  if (x <= 0 && y <= 0) {
    if (m <= b) {
      cout << "0" << endl;
      return;
    } else {
      cout << "-1" << endl;
      return;
    }
  }
  if (a < 0 && b > 0) {
    long long num = (-a) / b;
    a += b * num;
    ans += num;
  }
  if (max(b, a) >= m) {
    cout << "0" << endl;
    return;
  }
  for (long long i = 2;; i++) {
    long long c = a + b;
    if (c >= m) {
      cout << i - 1 + ans << endl;
      return;
    }
    a = min(b, c);
    b = max(b, c);
  }
}
int main() {
  cin >> x >> y >> m;
  solve();
  return 0;
}
