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 sigma(long long n) { return (n * (n + 1) / 2); }
bool great(pair<long long, long long> a, pair<long long, long long> b) {
  if (a.first != b.first) return a.first >= b.first;
  return a.second >= b.second;
}
int main() {
  long long a, b, x;
  cin >> a >> b >> x;
  vector<vector<long long>> ans;
  ans.resize(x + 1);
  long long one = 0;
  long long zero = 0;
  if (a > b) {
    for (long long i = 0; i < x + 1; i++) {
      if (i % 2) {
        ans[i].push_back(1);
        one++;
      } else {
        ans[i].push_back(0);
        zero++;
      }
    }
    long long rz = a - zero;
    long long ro = b - one;
    for (long long i = 0; i < rz; i++) {
      ans[0].push_back(0);
    }
    for (long long i = 0; i < ro; i++) {
      ans[1].push_back(1);
    }
  } else {
    for (long long i = 0; i < x + 1; i++) {
      if (i % 2) {
        ans[i].push_back(0);
        zero++;
      } else {
        ans[i].push_back(1);
        one++;
      }
    }
    long long rz = a - zero;
    long long ro = b - one;
    for (long long i = 0; i < rz; i++) {
      ans[1].push_back(0);
    }
    for (long long i = 0; i < ro; i++) {
      ans[0].push_back(1);
    }
  }
  for (long long i = 0; i < x + 1; i++) {
    for (long long j = 0; j < ans[i].size(); j++) {
      cout << ans[i][j];
    }
  }
}
