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;
const int INF = 1e9;
const int N = 1e5;
const int MOD = 1e9 + 7;
const double PI = acos(-1);
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  int a, b, x;
  cin >> a >> b >> x;
  string ans = "";
  x++;
  char cur = a > b ? '0' : '1';
  while (x) {
    ans += cur;
    a -= cur == '0';
    b -= cur == '1';
    cur = '0' + '1' - cur;
    x--;
  }
  bool f1 = 1, f2 = 1;
  for (int i = 0; i < ans.size(); i++) {
    if (ans[i] == '0' && f1) {
      for (int j = 0; j < a + 1; j++) cout << '0';
      f1 = 0;
    } else if (ans[i] == '1' && f2) {
      for (int j = 0; j < b + 1; j++) cout << '1';
      f2 = 0;
    } else
      cout << ans[i];
  }
}
