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. 
b, a, c = map(int, input().split())
bb = b
aa = a
cc = c
k = ''


def rec(s, a, b, c, k):
    if a == 0 and b == 0:
        if c != 0:
            return 'a'
        return k
    if c == 0:
        if s == '0':
            return rec('0', a, b - 1, c, k + '0')
        else:
            return rec('1', a - 1, b, c, k + '1')
    if c == 1:
        if s == '0':
            if b > 0:
                return rec(s, a, b - 1, c, k + s)
            else:
                return rec('1', a - 1, b, c - 1, k + '1')
        else:
            if a > 0:
                return rec(s, a - 1, b, c, k + s)
            else:
                return rec('0', a, b - 1, c - 1, k + '0')
    if s == '0':
        return rec('1', a - 1, b, c - 1, k + '1')
    else:
        return rec('0', a, b - 1, c - 1, k + '0')


d = rec('0', a, b - 1, c, '0')
if d.count('a') != 0:
    d = rec('1', a - 1, b, c, '1')
print(d)
