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.
import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.util.StringTokenizer;

public class C {
	
	private static StringTokenizer tokenizer;
    private static BufferedReader bf;
    private static PrintWriter out;
    
	private static int nextInt() throws IOException {
    	return Integer.parseInt(nextToken());
    }
    
	private static long nextLong() throws IOException {
    	return Long.parseLong(nextToken());
    }
    
    private static String nextToken() throws IOException {
    	while(tokenizer == null || !tokenizer.hasMoreTokens()) {
    		tokenizer = new StringTokenizer(bf.readLine());
    	}
    	return tokenizer.nextToken();
    }
    
	public static void main(String[] args) throws IOException {
		bf = new BufferedReader(new InputStreamReader(System.in));
    	tokenizer = null;
    	out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
    	long x = nextLong(); long y = nextLong();
    	long m = nextLong();
    	if(x >= m || y >= m) {
    		out.println(0);
    	}
    	else {
    		if(x <=0 && y <= 0)
    			out.println(-1);
    		else {
    			long[] f = new long[90];
    			f[0] = 1; f[1] = 1;
    			for(int i = 2; i < 90; i++)
    				f[i] = f[i-1] + f[i-2];
    			BigInteger res = BigInteger.ZERO;
    			long tx = Math.max(x, y);
    			long ty = Math.min(x, y);
    			if(ty < 0) {
    				if(Math.abs(ty) % tx == 0) {
    					res = res.add(BigInteger.valueOf(-ty/tx)); ty = 0;
    				}
    				else {
    					res = res.add(BigInteger.valueOf(-ty/tx + 1)); ty = ty + tx*((-ty)/tx+1);
    				}
    			}
    			int i;
    			for(i = 1; i < 90; i++) {
    				if(f[i]*tx + f[i-1]*ty >= m) {
    					res = res.add(BigInteger.valueOf(i)); break;
    				}
    			}
    			out.println(res);
    		}
    	}
    	out.close();
	}

}
