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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
public class Div3 {
    public static void main(String[] args) {
        FastReader ssc=new FastReader();
        int a= ssc.nextInt();
        int b= ssc.nextInt();
        int x= ssc.nextInt();
        boolean flag=false;
        if(a>b){
            flag=true;
        }

        String s="";
        int ones=0,zeros=0;
        for (int i = 0; i <a*b ; i++) {
            if(flag){

                flag=false;
                if(i==x-1){
                    for (int j = 0; j <a-zeros ; j++) {
                        s=s+0;
                    }
                    break;
                }else {
                    zeros++;
                    s = s + 0;
                }
            }else{

                flag=true;
                if(i==x-1){
                    for (int j = 0; j <b-ones ; j++) {
                        s=s+1;
                    }
                    break;
                }else {
                    s = s + 1;
                    ones++;
                }
        }
        }

        if(flag){
            for (int i = 0; i <a-zeros ; i++) {
                s=s+0;
            }
        }else
            for (int i = 0; i <b-ones ; i++) {
                s=s+1;
            }


        System.out.println(s);

    }
}


class FastReader{
    BufferedReader br;
    StringTokenizer st;
    public FastReader(){
        InputStreamReader inr=new InputStreamReader(System.in);
        br=new BufferedReader(inr);
    }


    String next(){
        while(st ==null || !st.hasMoreElements()){
            try{
                st=new StringTokenizer(br.readLine());
            }catch(IOException e){
                e.printStackTrace();
            }

        }
        return st.nextToken();

    }


    int nextInt(){
        return Integer.parseInt(next());
    }


    long nextLong(){
        return Long.parseLong(next());
    }




}

