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 -*-
"""
Created on Tue Jul  3 16:02:31 2018

@author: Arsanuos
"""

def main():
    rd = lambda: list(map(int, input().split()))
    a, b, x = rd()
    s = []
    o = ['1'] * b
    z = ['0'] * a
    flag = True if b > a else False
    
    for i in range(x + 1):
        if flag:
            s.append(o.pop())
        else:
            s.append(z.pop())
        flag = not flag
    if s[0] == '1':
        s = o + s
        o.clear()
    else:
        s = z + s
        z.clear()
    if s[-1] == '0':
        s += z
        z.clear()
    elif s[-1] == '1':
        s += o
        o.clear()
    
    if s[-2] == '0':
        for item in z:
            s.insert(len(s) - 2, item)
    if s[-2] == '1':
        for item in o:
            s.insert(len(s) - 2, item)
    
    for c in s:
        print(c, end="")
    print()
    
    
if __name__ == "__main__":
    main()