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;
template <typename F, typename S>
ostream &operator<<(ostream &os, const pair<F, S> &p) {
  return os << "(" << p.first << ", " << p.second << ")";
}
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
  os << "{";
  typename vector<T>::const_iterator it;
  for (it = v.begin(); it != v.end(); it++) {
    if (it != v.begin()) os << ", ";
    os << *it;
  }
  return os << "}";
}
template <typename T>
ostream &operator<<(ostream &os, const set<T> &v) {
  os << "[";
  typename set<T>::const_iterator it;
  for (it = v.begin(); it != v.end(); it++) {
    if (it != v.begin()) os << ", ";
    os << *it;
  }
  return os << "]";
}
template <typename F, typename S>
ostream &operator<<(ostream &os, const map<F, S> &v) {
  os << "[";
  typename map<F, S>::const_iterator it;
  for (it = v.begin(); it != v.end(); it++) {
    if (it != v.begin()) os << ", ";
    os << it->first << " = " << it->second;
  }
  return os << "]";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  ;
  int t = 1;
  while (t--) {
    int i, j, k;
    int one, zero, d;
    cin >> zero >> one >> d;
    if (d == 1) {
      string ans = "";
      while (zero--) {
        ans += "0";
      }
      while (one--) {
        ans += "1";
      }
      cout << ans << endl;
      return 0;
    }
    string ans;
    int now;
    if (one > zero) {
      ans = "10";
      now = 0;
    } else {
      ans = "01";
      now = 1;
    }
    one--;
    zero--;
    d--;
    string s[] = {"1", "0"};
    for (i = 1; i <= d - 1; i++) {
      ans += s[now];
      if (now == 0) {
        one--;
      } else
        zero--;
      now ^= 1;
    }
    if (ans[ans.size() - 1] == '0') {
      while (zero--) {
        ans += "0";
      }
      while (one--) {
        ans += "1";
      }
    } else {
      while (one--) {
        ans += "1";
      }
      while (zero--) {
        ans += "0";
      }
    }
    cout << ans << endl;
  }
}
