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;
const long long INF = 1e18;
int a, b, x;
string build(string p) {
  int pa = a, pb = b;
  string ret = p.substr(0, x);
  for (char c : ret) {
    if (c == '0')
      --pa;
    else
      --pb;
  }
  if (ret.back() == '0') {
    while (pa) {
      ret += '0';
      --pa;
    }
    while (pb) {
      ret += '1';
      --pb;
    }
  } else {
    while (pb) {
      ret += '1';
      --pb;
    }
    while (pa) {
      ret += '0';
      --pa;
    }
  }
  return ret;
}
bool check(string fin) {
  int ans = 0;
  for (int i = 0; i < fin.size() - 1; ++i) {
    if (fin[i] != fin[i + 1]) ++ans;
  }
  return ans == x;
}
int main() {
  ios::sync_with_stdio(0);
  cin >> a >> b >> x;
  string p = "";
  string q = "";
  for (int i = 0; i < 300; ++i) {
    p += "10";
    q += "01";
  }
  if (x == 1) {
    cout << string(a, '0') + string(b, '1') << endl;
  } else {
    string ans1 = build(p);
    string ans2 = build(q);
    if (check(ans1)) {
      cout << ans1;
    } else {
      cout << ans2;
    }
  }
  return 0;
}
