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. 
(zeros, ones, ireq) = input().split(' ')
(zeros, ones, ireq) = (int(zeros), int(ones), int(ireq))

out = ""

if zeros > ones:
    out = '0'
    zeros -= 1
else:
    out = '1'
    ones -= 1

i = 0
while i < ireq:
    if out[-1] == '0':
        out += '1'
        ones -= 1
    else:
        out += '0'
        zeros -= 1
    i += 1

lastOne = len(out) - 1
while not out[lastOne] == '1':
    lastOne -= 1

while ones > 0:
    out = out[:lastOne+1] + '1' + out[lastOne+1:]
    ones -= 1
 
lastZero = len(out) - 1
while not out[lastZero] == '0':
    lastZero -= 1
    
while zeros > 0:
    out = out[:lastZero+1] + '0' + out[lastZero+1:]
    zeros -= 1

print(out)