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;
using lli = long long int;
using pii = pair<int, int>;
using vi = vector<int>;
using vb = vector<bool>;
using vvi = vector<vector<int>>;
using vlli = vector<long long int>;
using vpii = vector<pair<int, int>>;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int a, b, x, idx = 0;
  cin >> a >> b >> x;
  string s;
  if (a > b)
    s = "0", a--;
  else
    s = "1", b--;
  while (x) {
    if (s[idx] == '0')
      s += '1', b--;
    else
      s += '0', a--;
    x--;
    idx++;
  }
  for (int i = 0; i < s.length(); i++) {
    cout << s[i];
    if (a && s[i] == '0')
      while (a) cout << '0', a--;
    if (b && s[i] == '1')
      while (b) cout << '1', b--;
  }
}
