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;
  cin >> a >> b >> x;
  int n = a + b;
  int p, q;
  int t = 0;
  string s = "";
  int what = 0;
  if (b > a) {
    what = 1;
    swap(a, b);
  }
  s = "0";
  a--;
  p = 1;
  q = 0;
  for (int i = 2; i <= x; i = i + 2) {
    s = s + "10";
    t = t + 2;
    p++;
    q++;
    a--;
    b--;
  }
  if (t != x) {
    s = s + "1";
    q++;
    b--;
  }
  if (a > 0) {
    string ns = "";
    for (int i = 1; i <= a; i++) {
      ns += "0";
    }
    s = ns + s;
  }
  if (b > 0) {
    string ns = "";
    for (int i = 1; i <= b; i++) {
      ns += "1";
    }
    if (s[s.length() - 1] == '1') {
      s = s + ns;
    } else {
      s = s.substr(0, s.length() - 1);
      s = s + ns + "0";
    }
  }
  if (what) {
    for (int i = 0; i < s.length(); i++) {
      if (s[i] == '0') {
        s[i] = '1';
      } else {
        s[i] = '0';
      }
    }
  }
  cout << s;
}
