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;
long long i, n, a, b, x;
vector<long long> m;
string s;
int main() {
  cin >> a >> b >> x;
  if (2 * a == x) {
    for (i = 0; i < a; i++) s += "01";
    for (i = a; i < b; i++) s = "1" + s;
    cout << s;
    return 0;
  }
  x++;
  long long l = 0;
  while (x > 0) {
    s += '0' + (l & 1);
    if (l & 1)
      b--;
    else
      a--;
    l++;
    x--;
  }
  while (a > 0) {
    s = '0' + s;
    a--;
  }
  while (b > 0) {
    if (s.back() == '0') {
      s.pop_back();
      s += "10";
    } else
      s += '1';
    b--;
  }
  cout << s;
  return 0;
}
