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.ArrayList;
import java.util.Arrays;
import java.util.Scanner;
import java.util.TreeSet;

public class Main {

    public static void main(String[] args) {
        Scanner input = new Scanner(System.in);

        int a[] = new int[2];
        a[0] = input.nextInt();
        a[1] = input.nextInt();
        int x = input.nextInt();

        int index = (a[0] > a[1]) ? 0 : 1;
        System.out.print(index);
        a[index]--;
        while (x-- > 1) {
            
            
            index = (++index) % 2;
            System.out.print(index);
            a[index]--;

        }
        while (a[index]-- > 0) {
            System.out.print(index);
        }
        index = ++index % 2;
        while (a[index]-- > 0) {
            System.out.print(index);
        }
        System.out.println("");

    }

}
