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.util.*;
import java.io.*;

public class PerfectPair {

	// https://codeforces.com/contest/318/problem/C
	
	public static void main(String[] args) throws IOException, FileNotFoundException {
		BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
		//BufferedReader in = new BufferedReader(new FileReader("PerfectPair"));

		StringTokenizer st = new StringTokenizer(in.readLine());
		long x = Long.parseLong(st.nextToken());
		long y = Long.parseLong(st.nextToken());
		long m = Long.parseLong(st.nextToken());
		if (x >= m || y >= m) {
			System.out.println(0);
			return;
		}
		if (m <= 0) {
			System.out.println(-1);
			return;
		}
		// m > 0
		if (x <= 0 && y <= 0) {
			System.out.println(-1);
			return;
		}
		/*if (x < 0 && Math.abs(x) > Math.abs(y)) {
			System.out.println(-1);
			return;
		}
		if (y < 0 && Math.abs(x) < Math.abs(y)) {
			System.out.println(-1);
			return;
		}*/
		
		if (y < x) {
			long temp = x;
			x = y;
			y = temp;
		}
		// m>0, x<=y
		
		long amount=0;
		if (x < 0) {
			// x + yk >= 0
			// k >= -x/y
			long k = (long) Math.ceil(-x/(double)y);
			x += y*k;
			amount += k;
		}
		
		long[] fib = new long[90];
		fib[0] = 1;
		fib[1] = 1;
		for (int i=2; i<fib.length; i++) {
			fib[i] = fib[i-1] + fib[i-2];
			//System.out.println(fib[i]);
		}
		
		if (y < x) {
			long temp = x;
			x = y;
			y = temp;
		}
		// m>0, x<=y
		
		if (y >= m) {
			System.out.println(amount);
			return;
		}
		
		int min=1;
		int max=fib.length;
		while (min < max) {
			int middle = (min+max)/2;
			if (fib[middle-1]*x + fib[middle]*y >= m) {
				max = middle;
			}
			else min = middle+1;
		}
		
		amount += min;
		System.out.println(amount);
	}
}