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 int MX = 1e6 + 5;
const int INF = 1e9;
const int M = 1000000007;
vector<int> adj[100005];
vector<int> vis(100005, 0);
int *input(int *n) {
  int v;
  cin >> v;
  *n = v;
  int *ar = (int *)malloc(sizeof(int) * v);
  for (int i = 0; i < v; i++) {
    cin >> ar[i];
  }
  return ar;
}
void solve() { return; }
int main() {
  ios_base::sync_with_stdio(false);
  int t = 1;
  while (t--) {
    int a, b, x;
    cin >> a >> b >> x;
    bool flg;
    if (a > b) {
      flg = 0;
    } else {
      flg = 1;
    }
    string str;
    if (flg == 0) {
      str = "0";
      a--;
    } else {
      str = "1";
      b--;
    }
    flg = !flg;
    for (int i = 0; i < x; i++) {
      if (flg == 0) {
        str = str + "0";
        a--;
      } else {
        str = str + "1";
        b--;
      }
      flg = !flg;
    }
    int n = str.size();
    for (int i = 0; i < n; i++) {
      if (str[i] == '0') {
        while (a > 0) {
          cout << 0;
          a--;
        }
      } else {
        while (b > 0) {
          cout << 1;
          b--;
        }
      }
      cout << str[i];
    }
  }
  return 0;
}
