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.*;
import java.lang.*;
import java.io.*;

/* Name of the class has to be "Main" only if the class is public. */
public class Codechef
{
	public static void main (String[] args) throws java.lang.Exception
	{
	 Scanner sc=new Scanner(System.in);
	 int a=sc.nextInt();
	 int b=sc.nextInt();
	 int x=sc.nextInt();
	 
	 int one=Math.max(a,b);
	 int zero=Math.min(a,b);
	 int req1=0;
	 String ans="";
	 
	 for(int i=0;i<x/2;i++)ans+="10";
	 one=one-x/2;zero-=x/2;
	 
	 if(x%2==1){
	 for(int i=0;i<one;i++)ans+=1;
	 for(int i=0;i<zero;i++)ans+=0;
	 }
	 else {
	   for(int i=0;i<zero;i++)ans+=0;
	   for(int i=0;i<one;i++)ans+=1;
	 }
	 char[] c=ans.toCharArray();
	 if(a>b){
	  for(int i=0;i<c.length;i++){
	    if(c[i]=='0')c[i]='1';
	    else c[i]='0';
	  }   
	 }
	 for(int i=0;i<c.length;i++)System.out.print(c[i]);
	}
}
