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. 
import sys

a, b, x = (map(int, sys.stdin.readline().split()))
out = ""
diff = ""
if x% 2 == 1:
    i = "0"
    for loop in range(x+1):
        diff += i
        if i == "1":
            i = "0"
        else:
            i = "1"
    for loop in range(a-(x+1)//2):
        out += "0"
    out += diff
    for loop in range(b-(x+1)//2):
        out += "1"
if x% 2 == 0:
    if a > b:
        diff = "01" + "1"*(b-x//2) + "01" *(x//2-1)+ "0" + "0"*((a-x//2)-1)
        out = diff
    else:
        diff = "10" + "0"*(a-x//2) + ("10")*(x//2-1)+ "1" + "1"*((b-x//2)-1)
        out = diff
print(out)
    


        