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;
import java.util.StringTokenizer;
import java.util.TreeMap;

/**
 * Created by ishaykhiev2017 on 03.07.2018.
 */
public class Rukzak {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int a = sc.nextInt();
        int b = sc.nextInt();
        int x = sc.nextInt();
        String s = "";
        int type = 0;
        int col1 = 0;
        int col0 = 0;
        for (int i = 0; i < x + 1; i++) {
            s += type;
            type += 1;
            type %= 2;
            if (type == 1) {
                col0++;
            } else {
                col1++;
            }
        }
        if (col1 > b||col0 > a) {
            String s2 = "";
            for (int i = 0; i < s.length(); i++) {
                if(s.charAt(i)=='1'){
                    s2+='0';
                }else{
                    s2+='1';
                }
            }
            s = s2;
            int tmp = col0;
            col0 = col1;
            col1 = tmp;
            for (int i = 0; i < b - col1 + 1; i++) {
                System.out.print(1);
            }
            for (int i = 0; i < a - col0 + 1; i++) {
                System.out.print(0);
            }
        }else {
            for (int i = 0; i < a - col0 + 1; i++) {
                System.out.print(0);
            }
            for (int i = 0; i < b - col1 + 1; i++) {
                System.out.print(1);
            }
        }
        System.out.println(s.substring(2, s.length()));
    }
}
