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())
s = "01010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010"
if x%2==1:
    t = s[:x+1]
    u = "0"*(a-t.count("0"))
    v = "1"*(b-t.count("1"))
    print(u+t+v)
else:
    if x==2:
        if a==1:
            print("1"+"0"+"1"*(b-1))
        elif b==1:
            print("0"+"1"+"0"*(a-1))
        else:
            print("0"+"1"*b+"0"*(a-1))
    else:
        if a >= b:
            t = s[:x-1]
            u = "0"*(a-t.count("0"))
            v = "1"*(b-t.count("1"))
            print(t+v+u)
        else:
            t = s[1:x]
            u = "0"*(a-t.count("0"))
            v = "1"*(b-t.count("1"))
            print(t+u+v)