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.util.StringTokenizer;

public class TC
{
    
    static long a, b, m;
    
    static long solve()
    {
        // both -ve
        
        if( ( a <= 0 ) && ( b<= 0 ) )
        {
            if( ( a >= m ) || ( b>=m ) )
                return 0;
            else
                return -1;
        }
        
         // ensure a<b
        
        if( a > b )
        {
            long t=a;
            a=b;
            b=t;
        }
        
        // -ve, +ve
         long tc=0;
         if( a< 0 )
         {
             tc=-a/b;
             a=a+tc*b;
         }
        
        // both +ve
              
        long i_2=b, i_1=a+b;
        if( i_2 >= m )
            return 0;
            if( i_1 >= m )
            return 1;
        long cnt=1+tc;
        
        while( true )
        {
            long curr=i_2 + i_1;
            cnt++;
            // sign change due to overflow
            if( ( i_2 >= 0 ) && ( i_1 >= 0 ) && ( curr < 0 ) )
                return cnt;
            if( curr >= m )
                return cnt;
            i_2=Math.max( i_1, i_2 );
            i_1=curr;
           
        }
    }
    
    public static void main( String[] args ) throws IOException
    {
        BufferedReader br=new BufferedReader( new InputStreamReader( System.in ));
        String s=br.readLine();
        StringTokenizer tok=new StringTokenizer(s);
        a=Long.parseLong(tok.nextToken());
        b=Long.parseLong(tok.nextToken());
        m=Long.parseLong(tok.nextToken());
        long res=solve();
        System.out.println( res );
    }
}