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>
using namespace std;
int main() {
  int a, b, x;
  scanf("%d%d%d", &a, &b, &x);
  int f = 0;
  if (b > a) swap(a, b), f = 1;
  string s = "0";
  for (int i = 0; i < x - 1; i++) {
    if (i & 1 ^ 1)
      s += '1';
    else
      s += '0';
  }
  string zeros = string(a - (s.size() + 1) / 2, '0');
  string ones = string(b - s.size() / 2, '1');
  string ans;
  if (s.back() == '1' && b > s.size() / 2) {
    ans = '1' + zeros + s + ones;
    ans.pop_back();
  } else if (s.back() == '1') {
    for (auto& it : s) it = char('1' - it + '0');
    ans = zeros + s + ones;
  } else
    ans = zeros + s + ones;
  if (f)
    for (auto& it : ans) it = char('1' - it + '0');
  cout << ans << endl;
  return 0;
}
