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 INF = 2e9 + 7;
const long long INF64 = 2e18;
const int N = 1e5 + 7;
int a, b, x;
int main(void) {
  scanf("%d %d %d", &a, &b, &x);
  int past = -1;
  while (1) {
    if (x == 1) {
      if (past == 1) {
        while (b--) printf("1");
        while (a--) printf("0");
      } else {
        while (a--) printf("0");
        while (b--) printf("1");
      }
      return 0;
    }
    if (past == -1) {
      if (a > b)
        printf("0"), past = 0, a--;
      else
        printf("1"), past = 1, b--;
    } else if (past == 1)
      printf("0"), a--, past = 0, x--;
    else
      printf("1"), b--, past = 1, x--;
  }
}
