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.*;
public class Main{

     public static void main(String []args){
        Scanner sc = new Scanner(System.in);
        
        int a = sc.nextInt();
        int b = sc.nextInt();
        int x = sc.nextInt();
        StringBuilder str = new StringBuilder();
        
        if(a>=b){
            str.append(0);  a--;
            int flag = 0;
            
            while(x>0){
                if(flag == 0){
                    str.append(1);
                    flag = 1;   b--;
                }
                else if(flag == 1){
                    str.append(0);
                    flag = 0;   a--;
                }
                x--;
            }
            while(a>0){
                int pos = str.indexOf("0");
                str.insert(pos,'0');
                a--;
            }
            while(b>0){
                int pos = str.indexOf("1");
                str.insert(pos,'1');
                b--;
            }
        }
        else{
            str.append(1);  b--;
            int flag = 0;
            
            while(x>0){
                if(flag == 0){
                    str.append(0);
                    flag = 1;   a--;
                }
                else if(flag == 1){
                    str.append(1);
                    flag = 0;   b--;
                }
                x--;
            }
            while(a>0){
                int pos = str.indexOf("0");
                str.insert(pos,'0');
                a--;
            }
            while(b>0){
                int pos = str.indexOf("1");
                str.insert(pos,'1');
                b--;
            }
        }
        System.out.println(str);
     }
}