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 T494 {
	public static void main(String[] args) {
		Scanner sc=new Scanner(System.in);
		int a=sc.nextInt();
		int b=sc.nextInt();
		int x=sc.nextInt();
		String s="";

		for(int i=x;i>=1;i-=2){
			if(b>a){
				s+="10";
			}else{
				s+="01";
			}
			b--;a--;
		}
		
		if(b>a){
			s+=x%2==0?"1":"";
			if(x%2==0)b--;
		}else{
			s+=x%2==0?"0":"";
			if(x%2==0)a--;
		}
//		System.out.println(a+" "+s);
		for(int i=0;i<a;i++){
			s=s.substring(0,s.indexOf("0"))+"0"+s.substring(s.indexOf("0"));
		}
		for(int i=0;i<b;i++){
			s=s.substring(0,s.indexOf("1"))+"1"+s.substring(s.indexOf("1"));
		}
		System.out.println(s);
	}
}
