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())
out = [0]
a -= 1
for i in range(x):
    next_left = int(not out[0])
    next_right = int(not out[-1])

    if next_left == 0 and a > 0:
        out = [0] + out
        a -= 1
    elif next_right == 0 and a > 0:
        out.append(next_right)
        a -= 1
    elif next_left == 1 and b > 0:
        out = [1] + out
        b -= 1
    elif next_right == 1 and b > 0:
        out.append(1)
        b -= 1
 
if a > 0:
    last_0_pos = len(out) - 1
    while out[last_0_pos] != 0:
        last_0_pos -= 1
    out = out[:last_0_pos] + ([0] * a) + out[last_0_pos:]

if b > 0:
    last_1_pos = len(out) - 1
    while out[last_1_pos] != 1:
        last_1_pos -= 1
    out = out[:last_1_pos] + ([1] * b) + out[last_1_pos:]

print(''.join(map(str, out)))

