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. 

import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner cin = new Scanner(System.in);
        int a, b, x;
        a = cin.nextInt();
        b = cin.nextInt();
        x = cin.nextInt();
        int[] ans = new int[210];
        int[] k = new int[2];
        k[0] = a;
        k[1] = b;
        int index = (a > b) ? 0 : 1;
        ans[1] = index;
        --k[index];
        System.out.print(ans[1]);
        for (int i =  2; i <= a + b; ++i) {
            if (x > 1) {
                index = index ^ 1;
                ans[i] = index;
                --k[index];
                --x;
            } else {
                if (k[index] != 0) {
                    ans[i] = index;
                    --k[index];
                } else {
                    ans[i] = index^1;
                    --k[index^1];
                }
            }
            System.out.print(ans[i]);
        }
    }
}