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.*;
import java.io.*;


public class B1003{
	public static void main(String args[])throws IOException{
		Scanner sc=new Scanner(new BufferedReader(new InputStreamReader(System.in)));
		int a=sc.nextInt();
		int b=sc.nextInt();
		int x=sc.nextInt();
		String s="";
		PrintWriter pw=new PrintWriter(System.out);
		if(b>=a){
		s="1";
		int max=Math.max(a,b)-1;
		int min=Math.min(a,b);
		for(int i=0;i<x;i++){
			if(s.charAt(s.length()-1)=='1')
			{
				s+='0';
				min--;
			}
			else{
				s+='1';
				max--;
			}
		}
		for(int i=0;i<max;i++)
			s='1'+s;
		for(int i=0;i<min;i++)
			s=s.substring(0,s.length()-1)+'0'+s.charAt(s.length()-1);
		}
		else{
			s="0";
		int max=a-1;
		int min=b;
		for(int i=0;i<x;i++){
			if(s.charAt(s.length()-1)=='1')
			{
				s+='0';
				max--;
			}
			else{
				s+='1';
				min--;
			}
		}
		for(int i=0;i<max;i++)
			s='0'+s;
		for(int i=0;i<min;i++)
			s=s.substring(0,s.length()-1)+'1'+s.charAt(s.length()-1);
		}
		pw.print(s);
		pw.close();

	}
}