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. 
a,b,x = [int(x) for x in raw_input().split()]
s = ''
if x==0:
	while a:
		s = s+p
		a-=1
	while b:
		s = s+q
		b-=1
t = (x+1)/2
if a>b:
	s = '01'*t
	a-=(x+1)/2
	b-=(x+1)/2
else:
	s = '10'*t
	a-=(x+1)/2
	b-=(x+1)/2
if a>b:
	if(a-1)>=0:
		s = '0'*(a-1) + s + '1'*b
		b=0
		a =1
else:
	if(b-1)>=0:
		s = '1'*(b-1) + s + '0'*a
		a =0
		b=1
if x%2 == 1:
	if a or b:
		s = s[0]+s
if x%2==0:
	if a>b:
		s = s+ '0'
	else:
		s = s + '1'
print s


