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. 
# -*- coding: utf - 8 -*-
"""""""""""""""""""""""""""""""""""""""""""""
|    author: mr.math - Hakimov Rahimjon     |
|    e-mail: mr.math0777@gmail.com          |
|    created: 01.01.1980 03:32              |
"""""""""""""""""""""""""""""""""""""""""""""
# inp = open("input.txt", "r"); input = inp.readline; out = open("output.txt", "w"); print = out.write
TN = 1


# ===========================================


def solution():
    a, b, x = map(int, input().split())
    a = ["0" for i in range(a)]
    b = ["1" for i in range(b)]
    s = []
    if len(a)<len(b):
        cur = 0
    else:
        cur = 1
    f0 = 0
    f1 = 0
    for i in range(x):
        if not cur:
            b = b[1:]
            f0 = i+1
            s.append("1")
            cur = 1
        else:
            f1 = i+1
            cur = 0
            a = a[1:]
            s.append("0")
    if f0>f1:
        s = s[:f0+1]+a+s[f0+1:]
        s = s[:f1+1]+b+s[f1+1:]
    else:
        s = s[:f1+1]+b+s[f1+1:]
        s = s[:f0+1]+a+s[f0+1:]
    print("".join(s))



# ===========================================
while TN != 0:
    solution()
    TN -= 1
# ===========================================
# inp.close()
# out.close()
