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 = list(map(int, input().split(' ')))

s = list()
if x%2 == 0:
	s.append('0')
	a-=1
else :
	s.append('1')
	b-=1

while x>0 and a>0 and b>0:
	if s[-1] == '1':
		s.append('0')
		a-=1
	else:
		s.append('1')
		b-=1
	x-=1

if x:
	if s[0] == '1' and a>0:
		s.insert(0, '0')
		a-=1
		x-=1
	elif s[0] == '0' and b>0:
		s.insert(0, '1')
		b-=1
		x-=1

if x:
	if s[-1] == '1' and a>0:
		s.append('0')
		a-=1
		x-=1
	elif s[-1] == '0' and b>0:
		s.append('1')
		b-=1
		x-=1

y = s.index('1')
for i in range(b):
	s.insert(y+i, '1')
y = s.index('0')
for i in range(a):
	s.insert(y+i, '0')
print(''.join(s))