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.*;
public class Test3{
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int a,b,x,i,a1,b1,temp=0;
		a = sc.nextInt();
		b = sc.nextInt();
		x = sc.nextInt();
		a1=a;
		b1=b;
		boolean z=a>b;
		String s="";
		if(a>b){
			s+="0";
			a1--;
		}
		else{
			s+="1";
			b1--;
		}
		while(temp<x){
			if(z){
				s+="1";
				b1--;
			}
			else{
				s+="0";
				a1--;
			}
			z=!z;
			temp++;
		}
		String s1="",s2="";
		while(a1>0){
			s1+="0";
			a1--;
		}
		while(b1>0){
			s2+="1";
			b1--;
		}
		i = s.indexOf('0');
		s = s.substring(0,i)+s1+s.substring(i);

		i = s.indexOf('1');
		s = s.substring(0,i)+s2+s.substring(i);
		
		System.out.println(s);

	}
}