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 Binary_String {

	public static void main(String [] args){
		
		Scanner in = new Scanner(System.in);
		
		int a = in.nextInt();
		int b = in.nextInt();
		int x = in.nextInt();
		
		in.close();
		
		String answer = "";
		
		if (a > b){
			answer += '0';
			a--;
		} else {
			answer += '1';
			b--;
		}
		
		while(x > 0){
			
			if (answer.endsWith("0")){
				answer += 1;
				b--;
			} else {
				answer += 0;
				a--;
			}
			
			x--;
			
		}
		
		
		String zeros = "";
		
		while(a-- > 0)
			zeros += 0;
		
		String ones = "";
		
		while(b-- > 0)
			ones += 1;
		
		for (int i = 0; i < answer.length(); i++)
			if (answer.charAt(i) == '0'){
				answer = answer.substring(0, i + 1) + zeros + answer.substring(i + 1, answer.length());
				break;
			}
		
		for (int j = 0; j < answer.length(); j++)
			if (answer.charAt(j) == '1'){
				answer = answer.substring(0, j + 1) + ones + answer.substring(j + 1, answer.length());
				break;
			}
		
		System.out.print(answer);
	}

}