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;
int Get() {
  char c;
  while (c = getchar(), (c < '0' || c > '9') && (c != '-'))
    ;
  bool Flag = (c == '-');
  if (Flag) c = getchar();
  int X = 0;
  while (c >= '0' && c <= '9') {
    X = X * 10 + c - 48;
    c = getchar();
  }
  return Flag ? -X : X;
}
void Output(int X) {
  if (X < 0) {
    putchar('-');
    X = -X;
  }
  int Len = 0, Data[10];
  while (X) {
    Data[Len++] = X % 10;
    X /= 10;
  }
  if (!Len) Data[Len++] = 0;
  while (Len--) putchar(Data[Len] + 48);
  putchar('\n');
}
int main() {
  long long X, Y, M;
  cin >> X >> Y >> M;
  if (X >= M || Y >= M) {
    cout << 0 << endl;
    return 0;
  }
  if (X < Y) swap(X, Y);
  if (X <= 0) {
    cout << -1 << endl;
    return 0;
  }
  long long Ans = 0;
  if (Y < 0) {
    Ans = -Y / X;
    Y += Ans * X;
  }
  while (X < M && Y < M) {
    if (X < Y)
      X += Y;
    else
      Y += X;
    Ans++;
  }
  cout << Ans << endl;
  return 0;
}
