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.
#_________________ Mukul Mohan Varshney _______________#

#Template
import sys
import os
import math
import copy
from math import gcd
from bisect import bisect
from io import BytesIO, IOBase
from math import sqrt,floor,factorial,gcd,log,ceil
from collections import deque,Counter,defaultdict
from itertools import permutations, combinations

#define function 
def Int(): return int(sys.stdin.readline())
def Mint(): return map(int,sys.stdin.readline().split())
def Lstr(): return list(sys.stdin.readline().strip())
def Str(): return sys.stdin.readline().strip()
def Mstr(): return map(str,sys.stdin.readline().strip().split())
def List(): return list(map(int,sys.stdin.readline().split()))
def Hash(): return dict()
def Mod(): return 1000000007
def Ncr(n,r,p): return ((fact[n])*((ifact[r]*ifact[n-r])%p))%p
def Most_frequent(list): return max(set(list), key = list.count)
def Mat2x2(n): return [List() for _ in range(n)]
def btod(n): 
    return int(n,2) 
    
def dtob(n): 
    return bin(n).replace("0b","")    

 
# Driver Code 	
def solution():
     #for _ in range(Int()):
          x,y,m=Mint()
          ans=0
          if(x>=m or y>=m):
               print(0)
          elif(x<=0 and y<=0):
               print(-1)
          else:
               if(x>0 and y<0):
                    ans=(x-y-1)//x
                    y+=ans*x
               elif(y>0 and x<0):
                    ans=(y-x-1)//y
                    x+=ans*y 
               while(x<m and y<m):
                    t=x+y
                    if(x<y):
                         x=t
                    else:
                         y=t
                    ans+=1     
               print(ans)          
               
          
          
#Call the solve function          
if __name__ == "__main__":
     solution()  