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 C {
    public static void main(String[] args){
        Scanner scanner=new Scanner(System.in);
        int n=scanner.nextInt(),m=scanner.nextInt(),k=scanner.nextInt();
        StringBuilder s=new StringBuilder();
        boolean /*lb=k<n&&k<m*/rb=n<m;
        String a=rb?"10":"01",z="",o="",ans="";
        for(int i=1;i<=k;i+=2){
            s.append(a);
            n--;
            m--;
        }
        for(int i=0;i<n;i++)z+="0";
        for(int i=0;i<m;i++)o+="1";
        if(k%2==0){
//            50 47 18
//            String t1="",t2="";
//            if(n==0||m==0){
//                if(s.charAt(0)=='0')ans=z+s+o+"L";
//                else ans=o+s+z;
//            }else{
                if(s.charAt(0)=='0') ans=z.substring(1)+s+o+"0";
                    //z.sub0 s+o+0
                else ans=o.substring(1)+s+z+"1";
//            }
        }else{
            if(s.charAt(0)=='1'){
                ans=o+s+z;
            }else{
                ans=z+s+o;
            }
        }
        System.out.println(ans);
    }
}
