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 main {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner s=new Scanner(System.in);
		int a=s.nextInt();//for zeroes
		int b=s.nextInt();//for ones
		int x=s.nextInt();//for indicies
		String str="";
		if(a>b)
		{
			str=str+'0';
			a--;
		}
		else 
		{
			str=str+'1';
			b--;
		}
		x--;
		while(a!=0 && b!=0 && x!=0)
		{
			char lc=str.charAt(str.length()-1);
			if(lc=='1')
			{
				str=str+'0';
				a--;
			}
			else 
			{
				str=str+'1';
				b--;
			}
			x--;
		}
		char lc=str.charAt(str.length()-1);
		if(lc=='0')
		{
			for(int i=0;i<a;i++)
			{
				str=str+'0';
			}
			for(int i=0;i<b;i++)
			{
				str=str+'1';
			}
		}
		else 
		{
			for(int i=0;i<b;i++)
			{
				str=str+'1';
			}
			for(int i=0;i<a;i++)
			{
				str=str+'0';
			}
		}
		
		System.out.println(str);
	}

}
