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 sys

def fiboo(big, small, m):
    cnt = 0
    while big+small < m:
        #big, small = max(big+small, big), min(big+small,big)
        big, small = big+small, big
        cnt += 1
        #print str(cnt) + ": " + str(big) + "/" + str(small)
    if big==m:
        return cnt
    else:
        return cnt + 1
    
def solve(x, y, m):
    if x>=m or y>=m:
        return 0
    if (x<0 or y<0) and (x>0 or y>0):
        sss = abs(min(x,y))
        sss2 = max(x,y)
        cnt = sss/sss2
        if sss % sss2 != 0:
            cnt += 1
        #print cnt
        return cnt + fiboo(sss2, -sss+cnt*sss2, m)
    if x>0 or y>0:
        return fiboo(max(x,y), min(x,y), m)

    return -1

x, y, m = map(int, sys.stdin.readline().split())

print int(solve(x,y,m))
