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. 
arr = [int(x) for x in input().split()]
ss = ""
x = arr[2]
start = True
if arr[0] > arr[1]:
    ss += "0"
else:
    ss += "1"
    start = False
i = 0
while x > 0:
    y = (int(ss[i]) + 1) % 2
    ss += str(y)
    x -= 1
    i += 1

num_z = arr[0] - int((arr[2] + 1) / 2)
num_o = arr[1] - int((arr[2] + 1) / 2)
if arr[2] % 2 == 0:
    if start:
        num_z -= 1
    else:
        num_o -= 1
#print(int((arr[2] + 1) / 2))
#print(num_z)
#print(num_o)
s1 = ""
s2 = ""
while num_z > 0:
    s1 += "0"
    num_z -=1
while num_o > 0:
    s2 += "1"
    num_o -= 1

def construct_str(ss, s1, s2):
    ss = s1 + ss
    if ss[0] == ss[len(ss) - 1]:
        ss = ss[0:len(ss)-1] + s2 + ss[len(ss)-1]
    else:
        ss += s2
    return ss;

if start:
    ss = construct_str(ss, s1, s2)
else:
    ss = construct_str(ss, s2, s1)
print(ss)