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())
df = a
kl = b
n = a + b
if a >= b:
    count = 0

    s = '0'

else:
    s = '1'
    count = 0

ca = 0
cb = 0
i = 0
while count != x:
    i += 1
    if s[i - 1] == '0':
        s += '1'

        count += 1
    else:

        s += '0'
        count += 1

a -= s.count('0')
b -= s.count('1')
sta = '0' * a
stb = '1' * b

if df >= kl:
    s = sta + s
    z = s.find('1')
    zk = s.find('0')
    s = s[0:z + 1] + stb + s[z + 1:]
else:
    s = stb + s
    z = s.find('0')
    s = s[0:z + 1] + sta + s[z + 1:]
z = 0

#     if s[i] != s[i + 1]:
#         z += 1
# if z != x:
#     z = s[-1]
#     if z == '0':
#         z = '1'
#
#     else:
#         z = '0'
#     s = s[:-1]+z
print(s)
