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. 
# import sys
# sys.stdin=open('F:\\C\\Script\\input.txt','r')
# sys.stdout=open('F:\\C\\Script\\output.txt','w')
# sys.stdout.flush()

a , b , x = [int(i) for i in input().split()]
s = '01'
xx = x//2
d = {'0':1 , '1':1}
if a > b:
	s = '01'
	it = '0'
	for i in range(x - 1):
		s += it
		d[it] += 1
		if it == '0':
			it = '1'
		else:
			it = '0'
	if s[-1] == '0':
		print ('0'*(a - d['0']) + s[:-1] + '1'*(b-d['1']) + '0')
	else:
		print ('0'*(a - d['0']) + s + '1'*(b-d['1']))
else:
	s = '10'
	it = '1'
	for i in range(x - 1):
		s += it
		d[it] += 1
		if it == '0':
			it = '1'
		else:
			it = '0'
	if s[-1] == '1':
		print ('1'*(b - d['1']) + s[:-1] + '0'*(a-d['0']) + '1')
	else:
		print ('1'*(b - d['1']) + s + '0'*(a-d['0']))