#include <bits/stdc++.h>
using namespace std;
const int INF = 1 << 29;
const int SIZE = 1e5 + 10;
inline int two(int n) { return 1 << n; }
inline int test(int n, int b) { return (n >> b) & 1; }
inline void set_bit(int& n, int b) { n |= two(b); }
inline void unset_bit(int& n, int b) { n &= ~two(b); }
inline int last_bit(int n) { return n & (-n); }
inline int ones(int n) {
  int res = 0;
  while (n && ++res) n -= n & (-n);
  return res;
}
int main() {
  long long x, y, m;
  cin >> x >> y >> m;
  if (x > m || y > m) {
    return puts("0"), 0;
  }
  long long ans = 0;
  if (x > 0 && y > 0 && m <= 0) ans = -1;
  if (x <= 0 && y <= 0 && m > 0) ans = -1;
  if (x < 0 && y < 0 && m >= 0) ans = -1;
  if (x < 0 && y < 0 && m < 0) x = abs(x), y = abs(y), m = abs(m);
  if (m < x && m < y) ans = -1;
  if (ans == -1) {
    return puts("-1"), 0;
  }
  if (x > y) swap(x, y);
  if (x < 0 && y > 0 && m > 0) {
    ans = x / y;
    ans = abs(ans);
    long long tmp = y * ans;
    x = x + tmp;
    if (x != 0) x += y, ans++;
  } else if (x < 0 && y > 0 && m < 0) {
    ans = y / x;
    ans = abs(ans);
    y = y - (x * ans);
    if (y != 0) y += x, ans++;
  }
  while (x < m && y < m) {
    if (x < y) {
      x = x + y;
    } else {
      y = y + x;
    }
    ans++;
  }
  cout << ans << endl;
  return 0;
}

### Prompt

In cpp, your task is to solve the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int INF = 1 << 29;
const int SIZE = 1e5 + 10;
inline int two(int n) { return 1 << n; }
inline int test(int n, int b) { return (n >> b) & 1; }
inline void set_bit(int& n, int b) { n |= two(b); }
inline void unset_bit(int& n, int b) { n &= ~two(b); }
inline int last_bit(int n) { return n & (-n); }
inline int ones(int n) {
  int res = 0;
  while (n && ++res) n -= n & (-n);
  return res;
}
int main() {
  long long x, y, m;
  cin >> x >> y >> m;
  if (x > m || y > m) {
    return puts("0"), 0;
  }
  long long ans = 0;
  if (x > 0 && y > 0 && m <= 0) ans = -1;
  if (x <= 0 && y <= 0 && m > 0) ans = -1;
  if (x < 0 && y < 0 && m >= 0) ans = -1;
  if (x < 0 && y < 0 && m < 0) x = abs(x), y = abs(y), m = abs(m);
  if (m < x && m < y) ans = -1;
  if (ans == -1) {
    return puts("-1"), 0;
  }
  if (x > y) swap(x, y);
  if (x < 0 && y > 0 && m > 0) {
    ans = x / y;
    ans = abs(ans);
    long long tmp = y * ans;
    x = x + tmp;
    if (x != 0) x += y, ans++;
  } else if (x < 0 && y > 0 && m < 0) {
    ans = y / x;
    ans = abs(ans);
    y = y - (x * ans);
    if (y != 0) y += x, ans++;
  }
  while (x < m && y < m) {
    if (x < y) {
      x = x + y;
    } else {
      y = y + x;
    }
    ans++;
  }
  cout << ans << endl;
  return 0;
}

```