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.
a,b,m=map(int,raw_input().split())
num=max(a,b)
cnt=0
#print a+b
if num>m:
    print 0
    quit()
    
if (((a+b<=0 and a<=0 and b<=0))):
    if num<m:
        print -1
    else:
        print 0
    quit()

if a+b<0:
    #only one is negative
    if a<0:
        cnt=abs(a/b)+1
        a+=b*cnt
    elif b<0:
        cnt=abs(b/a)+1
        b+=a*cnt
while max(a,b)<m:
    if a<b:
        a+=b
    else:b+=a
    cnt+=1
print cnt