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 String getString(int a, int x, char aa, char bb, int n){
        char[] chs = new char[n];
        int i = 1;
        if(x % 2 == 1) {
            chs[0] = aa;
            i = 2;
            x--;
            a--;
        }

        while(x > 0){
            chs[i] = aa;
            x -= 2;
            a--;
            i += 2;
        }

        i--;
        while(a > 0){
            chs[i] = aa;
            a--;
            i++;
        }

        for(i = 0; i < chs.length; i++)
            if(chs[i] != aa)
                chs[i] = bb;

        return new String(chs);

    }

    public static void main(String[] arg) {
        Scanner scan = new Scanner(System.in);
        int a = scan.nextInt();
        int b = scan.nextInt();
        int x = scan.nextInt();

        Main main = new Main();

        if(a > b)
            System.out.println(main.getString(b, x, '1', '0', a + b));
        else
            System.out.println(main.getString(a, x, '0', '1', a + b));
    }
}