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.IOException;
import java.math.BigInteger;

public class CFC {

	private static final int IN_BUFFER_SIZE = 1 << 16;
	private static final int OUT_BUFFER_SIZE = 1 << 16;
	private byte[] input = new byte[IN_BUFFER_SIZE];
	private int ix = IN_BUFFER_SIZE;
	private int bytesRead = ix;
	private byte[] output = new byte[OUT_BUFFER_SIZE];
	private int ox = 0;

	private void readMore() {
		try {
			bytesRead = System.in.read(input, 0, IN_BUFFER_SIZE);
			if (bytesRead <= 0)
				throw new RuntimeException();
			ix = 0;
		} catch (IOException e) {
			throw new RuntimeException();
		}
	}

	private void flushOut() {
		System.out.write(output, 0, ox);
		ox = 0;
	}

	private void append(char c) {
		if (ox == OUT_BUFFER_SIZE)
			flushOut();
		output[ox++] = (byte) c;
	}

	private long nextLong() {
		skipSpaces();
		long ret = 0;
		if (ix == bytesRead) {
			readMore();
		}
		int sign = 1;
		if (input[ix] == '-') {
			sign = -1;
			ix++;
		}
		while (true) {
			if (ix == bytesRead) {
				try {
					readMore();
				} catch (RuntimeException e) {
					return ret;
				}
			}
			if (input[ix] < '0')
				break;
			ret *= 10;
			ret += input[ix++] - '0';
		}
		return sign * ret;
	}

	int cntSize;

	private void skipSpaces() {
		while (true) {
			if (ix == bytesRead) {
				readMore();
			}
			if (input[ix] > ' ')
				break;
			ix++;
		}
	}

	private char[] nn = new char[32];

	private void printLong(long n) {
		if (n == 0) {
			append('0');
		} else {
			if (n < 0) {
				append('-');
				n = -n;
			}
			int kk = 0;
			while (n > 0) {
				nn[kk++] = (char) (n % 10 + '0');
				n /= 10;
			}
			for (int i = kk - 1; i >= 0; i--) {
				append(nn[i]);
			}
		}
	}

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

	private void work() {

		long x = nextLong();
		long y = nextLong();
		long m = nextLong();
		if (m <= x || m <= y) {
			append('0');
			append('\n');
		} else {
			if (x <= 0 && y <= 0) {
				append('-');
				append('1');
				append('\n');
			} else {
				if (x > y) {
					long t = x;
					x = y;
					y = t;
				}
				long k = 0;
				if (x < 0 && -x > y) {
					k = (-x) / y;
					x += k * y;
				}
				while (x <= y) {
					x += y;
					k++;
				}

				BigInteger[] f = new BigInteger[89];
				f[0] = BigInteger.ZERO;
				f[1] = BigInteger.ONE;
				for (int i = 2; i < f.length; i++) {
					f[i] = f[i - 1].add(f[i - 2]);
				}
				BigInteger X = BigInteger.valueOf(x);
				BigInteger Y = BigInteger.valueOf(y);
				BigInteger M = BigInteger.valueOf(m);
				for (int i = 0; i < f.length - 1; i++) {
					if ((f[i + 1].multiply(X).add(f[i].multiply(Y)))
							.compareTo(M) >= 0) {
						k += i;
						break;
					}
				}
				printLong(k);
				append('\n');
			}
		}

		if (ox > 0)
			flushOut();
		System.out.close();
	}
}
