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

if a>=b:
    t = '0'    
    while x>2:
        t += '10'
        x -= 2
    s = ''
    for i in range(a-t.count('0')):
        s += '0'
    for i in range(b-t.count('1')):
        s += '1'
    if x==2:
        res = s+t 
    else:
        res = t+s                
else:
    t = '1'
    while x>2:
        t += '01'
        x -= 2        
    s = ''
    for i in range(a-t.count('0')):
        s += '0'
    for i in range(b-t.count('1')):
        s += '1'
    if x==1:
        res = s+t
    else:
        res = t+s 
        
print(res)                     
            
