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.
x,y,m = map(int,input().split())
a=max(x,y)
if a<=0 and m>a:
    print(-1)
else:
    b=max(x,y);a=min(x,y)
    a1=a;b1=b
    if b>=m:
        print(0)
        exit()    
    ans = abs(b-a)//b
    a+=(b)*ans
    ans1=100000000000000000000
    while max(a,b)<m:
        if a<=b:
            a=a+b
        else:
            b=a+b
        ans+=1
    if a1>0:    
        ans1 = ((m-b1)//a1)
        if (m-b1)%a1:
            ans1+=1
    ans=min(ans,ans1)
    print(ans)        
            




    
