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, c, j = 0, i = 0, num = 0;
  char ans[3000] = {};
  scanf("%d %d %d", &a, &b, &c);
  for (; i < a;) ans[i++] = '0';
  for (; i < a + b;) ans[i++] = '1';
  int pos = a;
  while (1) {
    int cot = 0;
    for (i = 0; i < a + b - 1; i++)
      if (ans[i] != ans[i + 1]) cot++;
    if (cot == c) break;
    if (cot == c - 1 && b * 2 != c && a * 2 != c) {
      ans[a + b - 1] = '0';
      ans[pos - 1] = '1';
      break;
    }
    for (i = 0; i <= num * 2; i++) {
      char tmp = ans[pos + i];
      ans[pos + i] = ans[pos - 1 + i];
      ans[pos - 1 + i] = tmp;
    }
    pos--;
    num++;
  }
  printf("%s\n", ans);
  return 0;
}
