You are given three integers a, b and x. Your task is to construct a binary string s of length n = a + b such that there are exactly a zeroes, exactly b ones and exactly x indices i (where 1 ≤ i < n) such that s_i ≠ s_{i + 1}. It is guaranteed that the answer always exists.

For example, for the string "01010" there are four indices i such that 1 ≤ i < n and s_i ≠ s_{i + 1} (i = 1, 2, 3, 4). For the string "111001" there are two such indices i (i = 3, 5).

Recall that binary string is a non-empty sequence of characters where each character is either 0 or 1.

Input

The first line of the input contains three integers a, b and x (1 ≤ a, b ≤ 100, 1 ≤ x < a + b).

Output

Print only one string s, where s is any binary string satisfying conditions described above. It is guaranteed that the answer always exists.

Examples

Input

2 2 1


Output

1100


Input

3 3 3


Output

101100


Input

5 3 6


Output

01010100

Note

All possible answers for the first example: 

  * 1100; 
  * 0011. 



All possible answers for the second example: 

  * 110100; 
  * 101100; 
  * 110010; 
  * 100110; 
  * 011001; 
  * 001101; 
  * 010011; 
  * 001011. 
from math import*
t = input().split()
a = int(t[0])
b = int(t[1])
x = int(t[2])


aa=str(0)*a
bb=str(1)*b
s = ''

if(x%2==0):
	d=x//2
	w=1

else:
	d=(x+1)//2
	w=0

if(a>b):
	if(w==1):
		f=a-d-1
	else:
		f=a-d
	e = b-d
	h=0
	if(f>0):
		s+='0'*(f+1)
		h=1
	if(e>0):
		s+='1'*(e+1)
		h+=2

	if(h>2 and d!=1):	
		s+='01'*(d-1)
	elif(h==0):
		s+='01'*d	
	elif(h==1):
		s+='1'
		if(d!=1):
			s+='01'*(d-1)	
	
	if(w==1):
		s+='0'	
elif(a==b):
	if(w==1):
		f=a-d

		s+='0'*f
		s+='1'*f
		s+='10'*d
	elif(w==0):
		f=a-d
		s+='0'*f
		s+='01'*d
		s+='1'*f	
elif(a<b):
	if(w==1):
		f=b-d-1
	else:
		f=b-d
	e = a-d
	h=0
	if(f>0):
		s+='1'*(f+1)
		h=1
	if(e>0):
		s+='0'*(e+1)
		h+=2

	if(h>2 and d!=1):	
		s+='10'*(d-1)
	elif(h==0):
		s+='10'*d	
	elif(h==1):
		s+='0'
		if(d!=1):
			s+='10'*(d-1)	
	
	if(w==1):
		s+='1'	
print(s)			






		




