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 sys import stdin, stdout
ti = lambda : stdin.readline().strip()
ma = lambda fxn, ti : map(fxn, ti.split())
ol = lambda arr : stdout.write(' '.join(str(i) for i in arr) + '\n')
os = lambda i : stdout.write(str(i) + '\n')
olws = lambda arr : stdout.write(''.join(str(i) for i in arr) + '\n')

higher = None
lower = None
highernumber = None
lowernumber = None
a,b,x = ma(int, ti())
if a < b:
	highernumber = '1'
	higher = b
	lowernumber = '0'
	lower = a
else:
	highernumber = '0'
	higher = a
	lowernumber = '1'
	lower = b
ans = ''
if x % 2 == 0:
	while x/2 >= 1:
		ans += highernumber + lowernumber
		x -= 2
		higher -= 1
		lower -= 1
	while lower != 0:
		ans += lowernumber
		lower -= 1
	while higher != 0:
		ans += highernumber
		higher -= 1
else:
	while x/2 >= 1:
		ans += highernumber + lowernumber
		x -= 2
		higher -= 1
		lower -= 1
	while higher != 0:
		ans += highernumber
		higher -= 1
	while lower != 0:
		ans += lowernumber
		lower -= 1
os(ans)