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;
bool chkZero(long long x, long long y, long long m) {
  if (x == 0) {
    if (max(x, y) >= m) {
      printf("0");
    } else {
      printf("-1");
    }
  } else if (y == 0) {
    if (x >= m) {
      printf("0");
    } else {
      printf("-1");
    }
  }
  return true;
}
int main() {
  long long x, y, m;
  long long count = 0;
  scanf("%I64d %I64d %I64d", &x, &y, &m);
  if (max(x, y) == 0 or ((x == y) and x == 0)) {
    if (max(x, y) >= m) {
      printf("0");
    } else {
      printf("-1");
    }
    return 0;
  }
  long long maxs = max(x, y);
  long long old = min(x, y);
  long long tmp = maxs;
  if (x >= 0 and y >= 0) {
    while (maxs < m) {
      maxs += old;
      old = tmp;
      tmp = maxs;
      count++;
    }
    printf("%I64d", count);
  } else if ((x < 0 or y < 0) and (x > 0 or y > 0)) {
    if (max(x, y) >= m) {
      printf("0");
      return 0;
    }
    long long abs = fabs(min(x, y));
    count += (abs + 1) / max(x, y);
    if (x == min(x, y)) {
      x += max(x, y) * count;
    } else {
      y += max(x, y) * count;
    }
    maxs = max(x, y);
    old = min(x, y);
    tmp = maxs;
    while (maxs < m) {
      maxs = max(old + tmp, tmp);
      old = min(old + tmp, tmp);
      tmp = maxs;
      count++;
    }
    printf("%I64d", count);
  } else if ((x < 0 and y < 0)) {
    if (maxs >= m) {
      printf("0");
    } else
      printf("-1");
  }
  return 0;
}
