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>
int main() {
  int n, m, l, o, p;
  int a[350];
  while (~scanf("%d %d %d", &n, &m, &l)) {
    o = n;
    p = m;
    for (int i = 0; i <= l; i++) {
      if (o > p) {
        if (i % 2 == 0) {
          a[i] = 0;
          n--;
        } else {
          a[i] = 1;
          m--;
        }
      } else {
        if (i % 2 == 0) {
          a[i] = 1;
          m--;
        } else {
          a[i] = 0;
          n--;
        }
      }
    }
    int w = m;
    int e = n;
    for (int i = 0; i <= l; i++) {
      printf("%d", a[i]);
      if (a[i] == 1 && m > 0) {
        for (int j = 0; j < w; j++) {
          printf("1");
          m--;
        }
      }
      if (a[i] == 0 && n > 0) {
        for (int j = 0; j < e; j++) {
          printf("0");
          n--;
        }
      }
    }
    printf("\n");
  }
  return 0;
}
