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.Vector;

public class Main {

    public static void main(String args[]) {

        Scanner sc = new Scanner(System.in);
        int a = sc.nextInt(),b = sc.nextInt(),x = sc.nextInt();
        x++;
        Vector<Integer> v = new Vector<>();
        if(a > b) {
            while (x >= 2) {
                v.add(0);
                v.add(1);
                x -= 2;
                a--;
                b--;
            }
            if (x == 0) {
                for (int i = 0; i < a; i++)
                    System.out.print(0);
            }
            for (int i : v)
                System.out.print(i);
            for (int i = 0; i < b; i++)
                System.out.print(1);
            if(x != 0) {
                for (int i = 0; i < a; i++)
                    System.out.print(0);
            }
        } else {
            while (x >= 2) {
                v.add(1);
                v.add(0);
                x -= 2;
                a--;
                b--;
            }
            if (x == 0) {
                for (int i = 0; i < b; i++)
                    System.out.print(1);
            }
            for (int i : v)
                System.out.print(i);
            for (int i = 0; i < a; i++)
                System.out.print(0);
            if (x != 0) {
                for (int i = 0; i < b; i++)
                    System.out.print(1);
            }
        }
    }
}
