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 collections,itertools

def helper(a,b,x,s,res):
    if x == 1:
        if s[-1] == '0' and b:
            s += '0'*a + '1'*b
        elif s[-1] == '1' and a:
            s += '1'*b + '0'*a
        res.append(s)

    if res != []: return

    if s[-1] == '0':
        helper(a,b-1,x-1,s+'1',res)
    elif s[-1] == '1':
        helper(a-1,b,x-1,s+'0',res)

while True:
    try:
        a,b,x = map(int,input().split())
        if a > b:
            s = '0'
            a -= 1
        else:
            s = '1'
            b -= 1
        res = []
        helper(a, b, x, s, res)
        print(res[0])
    except:
        break


