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. 
#include <bits/stdc++.h>
int main() {
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(0);
  std::cout.tie(0);
  std::cerr.tie(0);
  int n0, n1, x;
  std::cin >> n0 >> n1 >> x;
  std::string s = n1 > n0 ? "1" : "0";
  while (x--) {
    s.push_back('1' - s.back() + '0');
  }
  for (auto& it : s) {
    n0 -= (it == '0');
    n1 -= (it == '1');
  }
  assert(n0 >= 0 && n1 >= 0);
  int p1 = s.find('1');
  s = s.substr(0, p1) + std::string(n1, '1') + s.substr(p1);
  int p0 = s.find('0');
  s = s.substr(0, p0) + std::string(n0, '0') + s.substr(p0);
  std::cout << s;
  return 0;
}
