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

s = raw_input()
#s = sys.stdin.readline()

#heavy = []

s = s.split()

x = int(s[0])
y = int(s[1])
m = int(s[2])

oper = 0

if x>= m or y >=m:
    print '0'
elif x<=0 and y<=0:
    print '-1'
else:    
    oper = 1
    if y<0:
        t = y
        y = x
        x = t
    if x <0:
        oper = abs(x)//y +1
        x = x + y*(oper-1)
        #print "promezh oper, x, y:", oper, x, y
        
    while x+y < m:
        if x<=y:
            x = x+y
        else:
            y=x+y
        oper += 1
    print oper
    
    
