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 ll = long long;
int main() {
  int tc = 1;
  while (tc--) {
    ll a, b, c;
    cin >> a >> b >> c;
    string s = "";
    if (c % 2 == 1) {
      ll t = c / 2 + 1;
      for (int i = 0; i < t; i++) s += "10";
      ll ze = a - t, on = b - t;
      string m = "";
      string r = "";
      while (on--) m += '1';
      while (ze--) r += '0';
      s = m + s;
      s += r;
      cout << s << endl;
    } else {
      ll k = 1;
      c--;
      ll t = c / 2 + 1;
      for (int i = 0; i < t; i++) s += "10";
      ll ze = a - t, on = b - t;
      string m = "";
      string r = "";
      int fg = 0, mg = 0;
      if (on) {
        fg = 1;
        on--;
      }
      if (ze) {
        mg = 1;
        ze--;
      }
      while (on > 0 && on--) m += '1';
      while (ze > 0 && ze--) r += '0';
      s = m + s;
      s += r;
      if (fg == 1 && mg == 1)
        s = "01" + s;
      else if (mg == 0 && fg == 1) {
        s += "1";
      } else if (mg == 1 && fg == 0) {
        s = "0" + s;
      }
      cout << s << endl;
    }
  }
  return 0;
}
