#include <bits/stdc++.h>
using namespace std;
long long d, x, y;
void extendedEuclid(long long A, long long B) {
  if (B == 0) {
    d = A;
    x = 1;
    y = 0;
  } else {
    extendedEuclid(B, A % B);
    int temp = x;
    x = y;
    y = temp - (A / B) * y;
  }
}
long long power(long long x, long long y, long long p) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
map<long long, long long> par, val;
void ini() {
  for (long long i = 0; i < 500010; i++) {
    par[i] = i;
  }
}
long long root(long long x) {
  while (par[x] != x) {
    par[x] = par[par[x]];
    x = par[x];
  }
  return x;
}
void uni(long long x, long long y) {
  x = root(x);
  y = root(y);
  par[y] = par[x];
}
long long isPrime(long long n) {
  if (n == 0 || n == 1) return n;
  if (n % 2 == 0) return (long long)2;
  if (n % 3 == 0)
    return (long long)3;
  else {
    for (long long i = 5; i * i <= n; i += 6) {
      if (n % i == 0) return i;
      if (n % (i + 2) == 0) return (i + 2);
    }
    return n;
  }
}
const long long N = 10000010;
vector<bool> prime(N);
vector<long long> spf(N);
void sieve() {
  fill(prime.begin(), prime.end(), true);
  prime[1] = false;
  prime[0] = false;
  long long i, j;
  for (i = 2; i * i <= N; i++) {
    if (prime[i] == true) {
      spf[i] = i;
      for (j = i * i; j < N; j += i) {
        prime[j] = false;
        if (spf[j] == 0) spf[j] = i;
      }
    }
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long n, i, j, k, m;
  cin >> n >> m >> k;
  if (n >= k || m >= k) {
    return cout << 0, 0;
  }
  if (n <= 0 && m <= 0) return cout << -1, 0;
  long long ans = 0;
  if (n <= 0) {
    double c = (double)(-n) / (double)m;
    c += (double)1;
    n = n + (ceil(c) - 1) * m;
    ans += ceil(c) - 1;
  }
  if (m <= 0) {
    double c = (double)(-m) / (double)n;
    c += (double)1;
    m = m + (ceil(c) - 1) * n;
    ans += ceil(c) - 1;
  }
  while (m < k && n < k) {
    ans++;
    if (min(m, n) == m)
      m = (m + n);
    else
      n = m + n;
  }
  cout << ans;
}

### Prompt

Your task is to create a CPP solution to 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;
long long d, x, y;
void extendedEuclid(long long A, long long B) {
  if (B == 0) {
    d = A;
    x = 1;
    y = 0;
  } else {
    extendedEuclid(B, A % B);
    int temp = x;
    x = y;
    y = temp - (A / B) * y;
  }
}
long long power(long long x, long long y, long long p) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
map<long long, long long> par, val;
void ini() {
  for (long long i = 0; i < 500010; i++) {
    par[i] = i;
  }
}
long long root(long long x) {
  while (par[x] != x) {
    par[x] = par[par[x]];
    x = par[x];
  }
  return x;
}
void uni(long long x, long long y) {
  x = root(x);
  y = root(y);
  par[y] = par[x];
}
long long isPrime(long long n) {
  if (n == 0 || n == 1) return n;
  if (n % 2 == 0) return (long long)2;
  if (n % 3 == 0)
    return (long long)3;
  else {
    for (long long i = 5; i * i <= n; i += 6) {
      if (n % i == 0) return i;
      if (n % (i + 2) == 0) return (i + 2);
    }
    return n;
  }
}
const long long N = 10000010;
vector<bool> prime(N);
vector<long long> spf(N);
void sieve() {
  fill(prime.begin(), prime.end(), true);
  prime[1] = false;
  prime[0] = false;
  long long i, j;
  for (i = 2; i * i <= N; i++) {
    if (prime[i] == true) {
      spf[i] = i;
      for (j = i * i; j < N; j += i) {
        prime[j] = false;
        if (spf[j] == 0) spf[j] = i;
      }
    }
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long n, i, j, k, m;
  cin >> n >> m >> k;
  if (n >= k || m >= k) {
    return cout << 0, 0;
  }
  if (n <= 0 && m <= 0) return cout << -1, 0;
  long long ans = 0;
  if (n <= 0) {
    double c = (double)(-n) / (double)m;
    c += (double)1;
    n = n + (ceil(c) - 1) * m;
    ans += ceil(c) - 1;
  }
  if (m <= 0) {
    double c = (double)(-m) / (double)n;
    c += (double)1;
    m = m + (ceil(c) - 1) * n;
    ans += ceil(c) - 1;
  }
  while (m < k && n < k) {
    ans++;
    if (min(m, n) == m)
      m = (m + n);
    else
      n = m + n;
  }
  cout << ans;
}

```