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.
//package codeforces.train;
import java.io.*;

public class ProblemA {
    BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
    PrintWriter writer = new PrintWriter(new OutputStreamWriter(System.out));

    int[] readInts() throws IOException {
        String[] strings = reader.readLine().split(" ");
        int[] ints = new int[strings.length];
        for(int i = 0; i < ints.length; i++) {
            ints[i] = Integer.parseInt(strings[i]);
        }
        return ints;
    }

    long[] ttl = null;
    int txl = 0;
    long readLong() throws IOException {
        if(ttl == null || txl >= ttl.length) {
            ttl = readLongs();
            txl = 0;
        }
        return ttl[txl++];
    }

    long[] readLongs() throws IOException {
        String[] strings = reader.readLine().split(" ");
        long[] longs = new long[strings.length];
        for(int i = 0; i < longs.length; i++) {
            longs[i] = Long.parseLong(strings[i]);
        }
        return longs;
    }

    int[] tt = null;
    int tx = 0;
    int readInt() throws IOException {
        if(tt == null || tx >= tt.length) {
            tt = readInts();
            tx = 0;
        }
        return tt[tx++];
    }

    void solve() throws IOException {
        long x = readLong();
        long y = readLong();
        long m = readLong();
        if(Math.max(x, y) >= m) {
            writer.println(0);
        }
        else if(x <= 0 && y <= 0) {
            writer.println(-1);

        }
        else {
            long ans = 0;
            while(Math.max(x, y) < m) {
                long min = Math.min(x, y);
                long max = Math.max(x, y);
                //max * t + min > max
                long t = max - min;
                t /= max;
                while(max * t + min <= max) t++;
                ans += t;
                x = max;
                y = max * t + min;
            }
            writer.println(ans);
        }
        writer.flush();
    }

    void multiSolve() throws IOException {
        int n = readInts()[0];
        for(int i = 0; i < n; i++) {
            if(i > 0) writer.println();
            solve();
        }
    }

    public static void main(String[] args) throws IOException{
        new ProblemA().solve();
    }
}
