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.Arrays;
import java.util.Scanner;

public class CodeTest {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int a=sc.nextInt();
		int b=sc.nextInt();
		int x=sc.nextInt();
		int count=0;
		String str="";
		int pass=0;
		while(count<x-1)
		{
			if(pass%2==0)
			{
				str+="1";
				b--;
			}
			else
			{
				str+="0";
				a--;
			}
			pass++;
			count=str.length()-1;	
		}
		//System.out.println(" count is "+count);
		if(str.endsWith("0"))
		{
			count++;
			if(count!=x||b==0)
			{
				str="0"+str;
				a--;
			}
			while(a-->0)
			{
				str+="0";
			}
			
			while(b-->0)
			{
				str+="1";
			}
 
		}
		else
		{
			while(b-->0)
				str+="1";
			while(a-->0)
				str+="0";
		}
		System.out.println(str);
	}
}