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())
mas = []
s = ''
for i in range(x+1):
    if i % 2 == 0:
        s += '1'
    else:
        s += '0'
mas.append(s)
s = list(s)
s.reverse()
mas.append(''.join(s))

s = ''
for i in range(x+1):
    if i % 2 == 0:
        s += '0'
    else:
        s += '1'
mas.append(s)
s = list(s)
s.reverse()
mas.append(''.join(s))

for el in mas:
    if el.count('1') <= b and el.count('0') <= a:
        s = el
a -= s.count('0')
b -= s.count('1')
s = list(s)
for i in range(x):
    if s[i] == '1':
        s[i] = '1' * (b + 1)
        break
s = list(''.join(s))
for i in range(len(s)):
    if s[i] == '0':
        s[i] = '0' * (a + 1)
        break
print(''.join(s))