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 long double PI = (long double)(3.1415926535897932384626433832795);
int parity_check(int n) {
  int par = n;
  par = par ^ (par >> 1);
  par = par ^ (par >> 2);
  par = par ^ (par >> 4);
  par = par ^ (par >> 8);
  par = par ^ (par >> 16);
  if (par % 2 == 1) {
    return 1;
  } else {
    return 0;
  }
}
int lcs(string X, string Y, int m, int n) {
  int L[m + 1][n + 1];
  int i, j;
  for (i = 0; i <= m; i++) {
    for (j = 0; j <= n; j++) {
      if (i == 0 || j == 0)
        L[i][j] = 0;
      else if (X[i - 1] == Y[j - 1] && i != j)
        L[i][j] = L[i - 1][j - 1] + 1;
      else
        L[i][j] = max(L[i - 1][j], L[i][j - 1]);
    }
  }
  return L[m][n];
}
int long long gcd(int long long a, int long long b) {
  if (a == 0) {
    return b;
  }
  return gcd(b % a, a);
}
int long long lcm(int long long a, int long long b) {
  int long long pro = a * b;
  int long long ans = gcd(a, b);
  pro = pro / ans;
  return pro;
}
int long long fast_power(int long long a, int long long b, int long long p) {
  if (b == 0) {
    return (1 % p);
  }
  int long long t = fast_power(a, b / 2, p);
  int long long ans = (t * t) % p;
  if (b % 2) {
    ans = (ans * (a % p)) % p;
  }
  return ans;
}
bool prime_check(int long long n) {
  int long long i, j;
  if (n == 1) {
    return false;
  }
  for (i = 2; i <= sqrt(n); i++) {
    if (n % i == 0) {
      return false;
    }
  }
  return true;
}
int long long fact(int long long n, int long long mod) {
  int long long prod = 1;
  for (int long long i = 1; i <= n; i++) {
    prod = (prod * i) % mod;
  }
  return prod;
}
void dfs(int long long s, int long long d,
         vector<pair<int long long, int long long> > v[], bool visited[],
         int long long *max, int long long *maxcount, int long long min,
         int long long count) {
  visited[s] = true;
  if (s == d) {
    if ((*max) < min) {
      (*max) = min;
      (*maxcount) = count;
    } else if ((*max) == min) {
      if ((*maxcount) < count) {
        (*maxcount) = count;
      }
    }
  } else {
    for (int long long i = 0; i < v[s].size(); i++) {
      if (!visited[v[s][i].first]) {
        if (min > v[s][i].second) {
          min = v[s][i].second;
        }
        dfs(v[s][i].first, d, v, visited, max, maxcount, min, count + 1);
      }
    }
  }
  visited[s] = false;
}
void polygonArea(int long long X[], int long long Y[], int long long n,
                 int long long dp[]) {
  int long long area = 0;
  int long long j = n - 1;
  for (int long long i = 0; i < n; i++) {
    area += (X[j] + X[i]) * (Y[j] - Y[i]);
    dp[i + 1] = area;
    j = i;
  }
}
void mera_sieve() {
  bool prime[6000013];
  memset(prime, true, sizeof(prime));
  for (int long long p = 2; p * p <= 6000012; p++) {
    if (prime[p] == true) {
      for (int long long i = p * p; i <= 6000012; i += p) prime[i] = false;
    }
  }
  for (int long long i = 2; i <= 6000012; i++) {
    if (prime[i]) {
    }
  }
}
string rev(string str) {
  string r = "";
  for (int i = str.length() - 1; i >= 0; i--) {
    r += str[i];
  }
  return r;
}
bool palindrome(string str) {
  int n = str.length();
  for (int i = 0; i < str.length(); i++) {
    if (str[i] != str[n - 1 - i]) {
      return false;
    }
  }
  return true;
}
int equal(vector<string> v1, vector<string> v2) {
  if (v1.size() != v2.size()) {
    return 0;
  }
  for (int i = 0; i < v1.size(); i++) {
    if (v1[i] != v2[i]) {
      return 0;
    }
  }
  return 1;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  int long long x, y, m;
  cin >> x >> y >> m;
  if (x >= m || y >= m) {
    cout << "0";
  } else if (x <= 0 && y <= 0) {
    cout << "-1";
  } else {
    int long long count = 0;
    if (x < 0 || y < 0) {
      if (x < 0) {
        count = ceil(abs(x) / (double)y);
        x = x + count * y;
      }
      if (y < 0) {
        count = ceil(abs(y) / (double)x);
        y = y + count * x;
      }
    }
    while (x < m && y < m) {
      int long long a = x + y;
      count++;
      if (x < y) {
        x = a;
      } else {
        y = a;
      }
    }
    cout << count;
  }
  return 0;
}
