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.IOException;
import java.util.Scanner;


public class Main{
	
	public static void main(String[] args) throws IOException{

		Scanner scanner = new Scanner(System.in);
		int a = scanner.nextInt();
		int b = scanner.nextInt();
		int x = scanner.nextInt();
		
		StringBuilder s = new StringBuilder();
		char c = (a > b)?'0':'1';
		for(int i = 0; i <= x; ++i) {
			s.append(c);
			if(c == '0') --a;
			else --b;
			
			c = (c == '0')?'1':'0';
		}
		
		
		for(int  i = 0; i < s.length() ; ++i) {
			if(s.charAt(i) == '0') {
				while(a-- > 0) s.insert(i, '0');
				break;
			}
		}
		
		
		for(int i = 0; i < s.length(); ++i) {
			if(s.charAt(i) == '1') {
				while(b-- > 0) {
					s.insert(i, '1');
				}
				
				break;
			}
		}
		
		System.out.println(s);
				
	}
	
	
}


