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 = map(int,raw_input().split())

s=''
c1=0
co=0
store =x
if x%2==0:
	s='0'
	co+=1
else:
	s='1'
	c1+=1
while x>0:
	if x%2==0:
		s+='1'
		c1+=1
	else:
		s+='0'
		co+=1
	x-=1
rem1 = b-c1
remo=a-co
temp=''

res=''

if store%2==0:
	res+='0'*(remo+1)
	res+='1'*(rem1)
	res+=s[1:]
	if remo<0:
		nres=''
		c1=0
		co=0
		for i in s:
			if i=='0':
				nres+='1'
				c1+=1
			else:
				nres+='0'
				co+=0
		res=nres+'1'*(b-c1)
else:
	for i in range(rem1):
		res+='1'
	res+=s
	for i in range(remo):
		res+='0'
print res
