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. 
zeroes, ones, indices = map(int, input().split())

result = ''
count = {'0': 0, '1': 0}
opposite = {'1': '0', '0': '1'}
result = '0' if zeroes > ones else '1'
count[result[0]] += 1

while indices > 0:
    count[opposite[result[len(result)-1]]] += 1
    result += opposite[result[len(result)-1]]
    indices -= 1

i = 0
while result[i] != '1':
    i += 1
result = result[:i] + '1' * (ones - count['1']) + result[i:]

i = 0
while result[i] != '0':
    i += 1
result = result[:i] + '0' * (zeroes - count['0']) + result[i:]


print(result)
