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?
Single line of the input contains three integers *x*, *y* and *m* ( - 10<sup class="upper-index">18</sup> ≤ *x*, *y*, *m* ≤ 10<sup class="upper-index">18</sup>).

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.
L=list(input().split())
for i in range(len(L)):L[i]=int(L[i])
m=L.pop()
ans=0
if sum(L)<=L[0] and sum(L)<=L[1]: ans=-1
else:
    while L[0]<m and L[1]<m:
        s=sum(L)
        if L[0]<L[1]: L[0]=s
        else: L[1]=s
        ans+=1
print(ans)
